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  \author[Suvrit Sra]{Suvrit Sra\textsuperscript{\normalfont*}}%
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    Email: \texttt{s.sra@tum.de}.}}

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\subjclass[2020]{Primary 05E05; Secondary 33C45, 26D07}
\keywords{Schur polynomial, homogeneous symmetric polynomial,
Tur\'an-ratio, Meixner, monotonicity}
\hypersetup{pdftitle={Monotonicity of Turan-ratios of symmetric polynomials},
pdfauthor={Suvrit Sra}}

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\begin{document}
\title[Tur\'an-ratios of symmetric polynomials]
{Monotonicity of Tur\'an-ratios of symmetric polynomials}
\aiauthor

\begin{abstract}
  We study the \emph{Tur\'an-ratio} $S_{\lambda,i}(x)^2/(S_{\lambda,i-1}(x)S_{\lambda,i+1}(x))$, where
  $S_{\lambda,i}(x):=s_\lambda(1^{[n-i]},x^{[i]})$, and $s_\lambda$ is the Schur polynomial in $n$
  variables (indexed by partition $\lambda$) with $i$ arguments equal to $x$ and $n-i$ of them equal to
  $1$. Specifically, we prove that this Tur\'an-ratio is strictly increasing for $x>1$ and $1\le i<n$
  (unless all $n$ padded parts of $\lambda$ are equal, then it is identically one). For one-row shapes,
  our result answers a question of Ait-Haddou about complete homogeneous symmetric polynomials. For
  this case, we also offer a second proof through the zeros of Meixner polynomials that gives an
  explicit positive formula for the logarithmic derivative. We conclude with a Tur\'an-ratio conjecture
  for Macdonald polynomials.
\end{abstract}
\preprintslink{sym-turan}{7 October 2026}
\maketitle

\vspace*{-8pt}
\section{Introduction}
Let $s_\lambda$ be the Schur polynomial in $n$ variables indexed by a partition $\lambda$ with at most
$n$ parts. Writing $a^{[k]}$ for $k$ repetitions, i.e., the \emph{multiplicity} of a scalar $a$, we set
\begin{equation}\label{eq:1}
  S_{\lambda,i}(x)=s_\lambda(1^{[n-i]},x^{[i]}),\quad0\le i\le n\quad
  \Tc_{\lambda,i}(x) :=
  \frac{S_{\lambda,i}(x)^2}
  {S_{\lambda,i-1}(x)S_{\lambda,i+1}(x)},\quad 1 \le i < n.
\end{equation}
If $\Tc_{\lambda,i}$ is strictly increasing for
$x>1$, then $S_{\lambda,i}(x)^2>S_{\lambda,i-1}(x)S_{\lambda,i+1}(x)$, an inequality of Tur\'an
type; we therefore call $\Tc_{\lambda,i}$ a \emph{Tur\'an-ratio}. Our main result is a proof of this monotonicity property.
\begin{theorem}[Schur multiplicity comparison]\label{thm:schur-turan}
  Let $n\ge2$ and let $\lambda$ be a partition with at most $n$ parts, padded with zeros to length
  $n$. If $\lambda_1>\lambda_n$, then $\Tc_{\lambda,i}$ is strictly increasing on $(1,\infty)$
  for every $1\le i<n$. If $\lambda=(r^n)$, all these ratios equal one.
\end{theorem}

The monotonicity question originates with a question asked by~\citet{haddou24}, who asked about the one-row  $\lambda=(\ell)$ version of~\eqref{eq:1}. More specifically, in this case $s_{(\ell)} = h_{\ell}$, where $h_{\ell}$ is the complete homogeneous symmetric polynomial of degree $\ell$:
\begin{equation}\label{eq:def-h}
 h_\ell(y)=\sum_{1\le j_1\le j_2\le\cdots\le j_\ell\le n}  y_{j_1}y_{j_2}\cdots y_{j_\ell}, \qquad h_0=1.
\end{equation}
For integers $\ell\ge1$ and $n\ge2$, we rewrite $H_i=S_{(\ell),i}$ and $\Psi_{i,n}=\Tc_{(\ell),i}$, so that~\eqref{eq:1} becomes
\begin{equation}\label{eq:Hi-def}
 H_i(x)=h_\ell(1^{[n-i]},x^{[i]}),\qquad \Psi_{i,n}(x)=\frac{H_i(x)^2}{H_{i-1}(x)H_{i+1}(x)}.
\end{equation}
\citet{haddou24} asked whether $\Psi_{i,n}$ is increasing on $(1,\infty)$ for $n\ge4$ and $2\le i\le n-2$, and reported a proof for $\Psi_{2,4}$. Taking $\lambda=(\ell)$ in Theorem~\ref{thm:schur-turan} resolves \citeauthor{haddou24}'s  question.

\begin{corr}\label{thm:haddou}
For integers $\ell\ge1$, $n\ge2$, and $1\le i< n$, $\Psi_{i,n}$ is strictly increasing on $(1,\infty)$.
\end{corr}

Section~\ref{sec:schur-toda} proves Theorem~\ref{thm:schur-turan} through a Toda identity for the
confluent determinants of \citet{gonzalez2023bialternant} and a maximum principle that treats all
multiplicities $i$ at once. For one-row shapes, Section~\ref{sec:meixner} gives an independent proof
through the zeros of Meixner polynomials, which also yields a positive formula for the
log-derivative of $\Psi_{i,n}$. Section~\ref{sec:macdonald-conjecture} \emph{conjectures} (with
some evidence) that the same monotonicity holds for Macdonald polynomials $P_\lambda(\,\cdot\,;q,t)$
when the repeated arguments are replaced by an aligned split of the principal specialization
$(1,t,\dots,t^{n-1})$.

\section{Proof of Theorem~\ref{thm:schur-turan}: Tur\'an-ratios for Schur polynomials}\label{sec:schur-toda}
The ratio $\Tc_{\lambda,i}$ compares three Schur values that differ only in the multiplicity of $x$. By the confluent bialternant formula for Schur polynomials, these values are determinants from one
family $(D_0,\ldots,D_n)$, and adjacent members of such a family satisfy a \emph{Toda identity}: a
second-order differential identity in $t=\log x$ that links $D_i$ to $D_{i\pm1}$. This identity provides the key structure that we exploit and turn into a maximum principle for the derivatives of $\log\Tc_{\lambda,i}(e^t)$.

\subsection{The Toda identity}
First, we make a change of coordinates $t=\log x$. Rewriting~\eqref{eq:1}, 
\[
 f_i(t) :=\log S_{\lambda,i}(e^t),\qquad
 u_i(t) := \log\Tc_{\lambda,i}(e^t) = 2f_i(t)-f_{i-1}(t)-f_{i+1}(t),
\]
our goal is to prove $u_i'(t)>0$ on $(0,\infty)$, but we have no direct handle on $u_i'$. What the
determinant structure of the $S_{\lambda,i}$ gives is a differential equation, and it is second order in
$f_i$: Lemma~\ref{lem:schur-toda} expresses $f_i''$ through $u_i$. Taking the second difference in $i$
of both sides then gives $u_i''$ in terms of $u_{i-1},u_i,u_{i+1}$, eliminating $f$; differentiating
once more then gives a linear system for the $u_i'$, whose sign Section~\ref{sec:maximum-principle}
controls from its boundary values. The structure of Schur polynomials is key to this
lemma. %

\begin{lemma}[Toda]\label{lem:schur-toda}
Let $n\ge2$; let $\lambda$ be a partition with at most $n$ parts; for $t>0$ and $1\le i< n$,
\begin{equation}\label{eq:schur-toda}
  f_i''(t)=\frac{i(n-i)}{4\sinh^2(t/2)}
  \bigl(1-e^{-u_i(t)}\bigr).
\end{equation}
\end{lemma}

\begin{proof}
  \textbf{\emph{Step 1}}. We first express $S_{\lambda,i}$ as a determinant. Pad
$\lambda$ with zeros and set $a_j=\lambda_{n-j}+j$ for $0\le j<n$, so that
$a_0<\cdots<a_{n-1}$. The \emph{confluent bialternant formula} of
\citet[Theorem~1.1 and (4.5)]{gonzalez2023bialternant} evaluates a Schur
polynomial at repeated arguments as a quotient of two determinants; for the
alphabet $(1^{[n-i]},x^{[i]})$ it reads
\[
 S_{\lambda,i}(x)=\det G_\lambda/\det G_\varnothing,
\]
where $G_\lambda$ has rows indexed by $j$ and columns
$\frac1{r!}\partial_y^{\,r}y^{a_j}$, at $y=1$ for $0\le r<n-i$ and at $y=x$ for
$0\le r<i$; and $G_\varnothing$ is the same matrix for $\lambda=\varnothing$, so
that $\det G_\varnothing=(x-1)^{i(n-i)}$ \citep[(4.6)]{gonzalez2023bialternant},
with sign $+$ for the row order $a_0<\cdots<a_{n-1}$.

The derivatives in $y$ are not that convenient. Step~2 needs a determinant
whose entries are successive derivatives in $t=\log x$ of one function, and $\partial_t=y\partial_y$ acts on $y^{a_j}=e^{a_jt}$ by multiplication with
$a_j$. We thus replace $\partial_y^{\,r}$ by $(y\partial_y)^r$: with the
row vector
$d_j=(1,a_j,\dots,a_j^{\,n-i-1},e^{a_jt},a_je^{a_jt},\dots,a_j^{\,i-1}e^{a_jt})$,
define $D_i(t):=\det[d_j]_{j=0}^{n-1}$, whose columns are $(y\partial_y)^ry^{a_j}$
at $y=1$ for $0\le r<n-i$ and at $y=x=e^t$ for $0\le r<i$. Since
$(y\partial_y)^r=y^r\partial_y^{\,r}+\sum_{k<r}c_{r,k}\,y^k\partial_y^{\,k}$ with
constants $c_{r,k}$, subtracting from each column of $D_i$ suitable multiples of
the lower-order columns of the same block leaves $y^r\partial_y^{\,r}y^{a_j}$,
which is $r!\,y^r$ times the corresponding column of $G_\lambda$. The product of
these scalars over both blocks gives
\[
 D_i(t)=C_i\,x^{i(i-1)/2}\det G_\lambda,\quad\text{where}\quad
 C_i=\left(\nlprod_{r=0}^{n-i-1}r!\right)\left(\nlprod_{r=0}^{i-1}r!\right),
\]
and the confluent bialternant formula turns this into
\begin{equation}\label{eq:schur-confluent}
 D_i(t)=C_i e^{i(n-1)t/2}[2\sinh(t/2)]^{i(n-i)}S_{\lambda,i}(e^t).
\end{equation}
For $i=0,n$, this formula follows from ordinary Vandermonde; in particular,
$D_i(t)>0$ for $t>0$. %

\vskip4pt
\noindent\textbf{\emph{Step 2}}. Identity~\eqref{eq:schur-toda} relates $f_i''$ to
$u_i=2f_i-f_{i-1}-f_{i+1}$, and by~\eqref{eq:schur-confluent} each $f_k$ is $\log D_k$ up to explicit
terms; so~\eqref{eq:schur-toda} is a relation between $D_i$ and its neighbors $D_{i\pm1}$. Hankel
determinants of successive derivatives of one function carry such a relation: for
$K_i=\det[g^{(r+s)}]_{r,s<i}$, the Desnanot--Jacobi identity gives
$K_iK_i''-(K_i')^2=K_{i-1}K_{i+1}$. We show that $D_i$ is such a determinant up to a constant, and
then read off~\eqref{eq:schur-toda}. Introduce
\[
 g(t)=\sum_{j=0}^{n-1}
       \frac{e^{a_jt}}{\prod_{k\ne j}(a_j-a_k)},\qquad
 K_i(t)=\det[g^{(r+s)}(t)]_{r,s=0}^{i-1},\qquad K_0=1.
\]
Cauchy--Binet expands $K_i$ over subsets of $i$ exponents, and
expanding $D_i$ by its last $i$ columns gives the same terms, with
\begin{equation}\label{eq:schur-hankel}
 K_i(t)=(-1)^{i(i-1)/2}\frac{D_i(t)}{\Delta(a)},\qquad
 \Delta(a)=\prod_{j<k}(a_k-a_j).
\end{equation}
In the Desnanot--Jacobi identity the two bordered minors are $K_i'$ and $K_i''$: differentiating
$K_i$ row by row, only the derivative of the last row survives, since differentiating any other row
duplicates the next one; differentiating that minor column by column likewise leaves only the last
column, so $K_i''$ is the minor with last row and last column shifted. By~\eqref{eq:schur-hankel},
$\tau_i=(-1)^{i(i-1)/2}K_i=D_i/\Delta(a)>0$, and the alternating signs give
\[
 (\log\tau_i)''=-\frac{\tau_{i-1}\tau_{i+1}}{\tau_i^2},\qquad 1\le i<n.
\]
By~\eqref{eq:schur-confluent}--\eqref{eq:schur-hankel} and
$C_{i-1}C_{i+1}/C_i^2=i(n-i)$, the right-hand side is
$-i(n-i)e^{-u_i}/[4\sinh^2(t/2)]$, and the left-hand side is
$f_i''-i(n-i)/[4\sinh^2(t/2)]$. This proves~\eqref{eq:schur-toda}.
\end{proof}

\subsection{A maximum principle for the derivatives}\label{sec:maximum-principle}

The proof of Theorem~\ref{thm:schur-turan} differentiates
\eqref{eq:schur-toda} once more and arrives at the
system~\eqref{eq:chain-system} for the derivatives $u_i'$. The next
lemma shows that components of such a system that are nonnegative at
both ends of $(0,\infty)$ are nonnegative throughout, and positive
unless they all vanish. The coupling in~\eqref{eq:chain-system} has the
right sign for the sum of the negative parts of the components to
satisfy a scalar differential inequality, and the ordinary maximum
principle then applies to that sum.

\begin{lemma}[Maximum principle]\label{lem:chain-maximum}
Let $n\ge2$, let $p$ and $b_1,\dots,b_{n-1}>0$ be continuous on
$(0,\infty)$, and let $v_1,\dots,v_{n-1}\in C^2(0,\infty)$ satisfy
\begin{equation}\label{eq:chain-system}
 v_i''+p v_i'
 =2b_iv_i-b_{i-1}v_{i-1}-b_{i+1}v_{i+1},\qquad 1\le i<n,
\end{equation}
where $b_0v_0=b_nv_n=0$. If each $v_i$ has nonnegative limits at $0$
and $\infty$, then $v_i\ge0$ for every $i$. If moreover $v_i(t_0)=0$
for some $i$ and some $t_0>0$, then $v_k\equiv0$ for every $k$.
\end{lemma}

\begin{proof}
We show that $w=\sum_{i=1}^{n-1}\max\{-v_i,0\}$ vanishes identically.
At a point where the nonzero signs are fixed let $I=\{i:v_i<0\}$.
Summing the equations over these indices gives
\[
 w''+p w'
 =-\sum_{j=1}^{n-1}b_jv_j
       \bigl(2\mathbf1_I(j)-\mathbf1_I(j-1)-\mathbf1_I(j+1)\bigr)
 \ge0.
\]
Here $\mathbf1_I(0)=\mathbf1_I(n)=0$. For $j\in I$, the factor in
parentheses is nonnegative and $v_j<0$; for $j\notin I$, the signs
are reversed. The positive-part chain rule preserves this inequality
distributionally: a crossing through zero contributes a nonnegative
jump to the derivative of the negative part.

With $P'=p$, this inequality reads $(e^Pw')'\ge0$, so
$w\le\max\{w(a),w(b)\}$ on every $[a,b]\subset(0,\infty)$. Since
$w\to0$ at $0$ and at $\infty$, letting $a\to0$ and $b\to\infty$
gives $w\equiv0$.

Suppose $v_i(t_0)=0$ with $t_0>0$. Since $v_i\ge0$, we have
$v_i'(t_0)=0$ and $v_i''(t_0)\ge0$, whereas~\eqref{eq:chain-system}
forces $v_{i\pm1}(t_0)=0$. Repeating in $i$ gives
$v_k(t_0)=v_k'(t_0)=0$ for every $k$, and uniqueness for the linear
system gives $v_k\equiv0$.
\end{proof}

\begin{proof}[\textbf{Proof of Theorem~\ref{thm:schur-turan}}]
We first bring Lemma~\ref{lem:schur-toda} into the
form~\eqref{eq:chain-system}. Write
\[
 \alpha_i=i(n-i),\qquad c(t)=\frac1{4\sinh^2(t/2)},\qquad
 b_i(t)=\alpha_i c(t)e^{-u_i(t)},\qquad v_i(t)=u_i'(t),
\]
and set $b_0=b_n=0$. Since
$2\alpha_i-\alpha_{i-1}-\alpha_{i+1}=2$, Lemma~\ref{lem:schur-toda}
gives
\[
 u_i''=2c-2b_i+b_{i-1}+b_{i+1}.
\]
For $p(t)=\coth(t/2)$, the derivatives are $c'=-pc$ and
$b_i'=-(p+v_i)b_i$. Differentiating the last display therefore cancels
the terms involving $c$ and gives exactly~\eqref{eq:chain-system}.

Lemma~\ref{lem:chain-maximum} needs the boundary values of $v_i$.
At $t=0$ the arguments coincide, and symmetry and homogeneity give
$f_i'(0)=i|\lambda|/n$, so $v_i(0)=0$.
At infinity, $\deg S_{\lambda,i}=\lambda_1+\cdots+\lambda_i=:d_i$.
To see this, order the alphabet so
that the $i$ copies of $x$ come first. A semistandard tableau then has
at most $i$ entries from these copies in each column, and filling row
$j$ with $j$ attains the bound. Since the leading coefficient is
positive,
\[
 \lim_{t\to\infty}v_i(t)
 =2d_i-d_{i-1}-d_{i+1}=\lambda_i-\lambda_{i+1}\ge0.
\]

Lemma~\ref{lem:chain-maximum} now gives $v_i\ge0$ for every $i$. If
$\lambda_1>\lambda_n$, then $\lambda_i>\lambda_{i+1}$ for some $i$, so
$v_i\not\equiv0$, and the lemma gives $v_k>0$ on $(0,\infty)$ for every
$k$; that is, each $\Tc_{\lambda,k}$ is strictly increasing on
$(1,\infty)$. For $\lambda=(r^n)$, the identity
$s_\lambda(y)=(y_1\cdots y_n)^r$ gives $\Tc_{\lambda,i}=1$
directly.
\end{proof}

\section{Tur\'an-ratios monotonicity for $h_\ell$ through Meixner zeros}\label{sec:meixner}

This section gives a second proof of Corollary~\ref{thm:haddou},
independent of Section~\ref{sec:schur-toda}, together with the
positive formula~\eqref{eq:logpsi-deriv} for $(\log\Psi_{i,n})'$. To factor the ratio, we express $H_i$ as a polynomial
evaluated at the negative multiplicity $-i$. The zeros of this
polynomial are positive and move monotonically with $x$, and factoring
$\Psi_{i,n}$ over them proves the corollary.

\subsection{The Meixner representation}

The multiplicity $i$ enters $H_i(x)=h_\ell(1^{[n-i]},x^{[i]})$ only
through the number of arguments equal to $x$, and the point of this
section is that the dependence on $i$ is polynomial, of a classical
kind: $H_i(x)$ is a Meixner polynomial evaluated at $-i$. For $x>1$ put
$c=1/x\in(0,1)$ and let
\begin{equation}\label{eq:meixner-def}
 M_\ell(u;n,c):={}_2F_1\left(-\ell,-u;n;1-\frac1c\right)
 =\sum_{r=0}^{\ell}\frac{(-\ell)_r(-u)_r}{(n)_r\,r!}\left(1-\frac1c\right)^{\!r}
\end{equation}
denote the Meixner polynomial of degree $\ell$ in the variable $u$, in
the normalization of \citet[\S9.10]{koekoek2010hypergeometric}. Here
$(a)_r=a(a+1)\cdots(a+r-1)$ is the rising factorial, with $(a)_0=1$.

\begin{lemma}[Meixner representation]\label{lem:meixner-rep}
For integers $0\le i\le n$ and $x>1$,
\begin{equation}\label{eq:H-meixner}
 H_i(x)=\binom{n+\ell-1}{\ell}M_\ell(-i;n,c),
 \qquad c=\frac1x.
\end{equation}
\end{lemma}

\begin{proof}
Extract the coefficient of $t^\ell$ from
\[
 \sum_{L\ge0}h_L\bigl(1^{[n-i]},x^{[i]}\bigr)t^L
 =(1-t)^{-(n-i)}(1-xt)^{-i}
 =(1-t)^{-n}\left(1-\frac{(x-1)t}{1-t}\right)^{\!-i}.
\]
Expanding the last factor and using $[t^{\ell-r}]\,(1-t)^{-n-r}=\binom{n+\ell-1}{\ell-r}$ gives
\[
 H_i(x)=\sum_{r=0}^{\ell}\frac{(i)_r}{r!}\binom{n+\ell-1}{\ell-r}(x-1)^r,
\]
and $\binom{n+\ell-1}{\ell-r}=\binom{n+\ell-1}{\ell}\binom{\ell}{r}\frac{r!}{(n)_r}$ turns this sum into
\begin{equation}\label{eq:h-expansion}
 \frac{H_i(x)}{\binom{n+\ell-1}{\ell}}
 =\sum_{r=0}^{\ell}\binom{\ell}{r}\frac{(i)_r}{(n)_r}(x-1)^r.
\end{equation}
Finally, $\frac{(-\ell)_r}{r!}(1-x)^r=\binom{\ell}{r}(x-1)^r$ and $(-(-i))_r=(i)_r$. Because $1-c^{-1}=1-x$, the right side of~\eqref{eq:h-expansion} is therefore $M_\ell(-i;n,c)$.
\end{proof}

\subsection{Orthogonality and positive zeros}

The proof of the corollary in Section~\ref{sec:factorization} factors
$M_\ell$ over its zeros, so \eqref{eq:H-meixner} reduces the problem
to the location of those zeros and to their dependence on $c$.
Orthogonality places them on the positive axis; differentiating
orthogonality in Section~\ref{sec:zero-motion} then determines how
they move with $c$.

\begin{lemma}[Orthogonality and positive zeros]\label{lem:meixner-orth}
Let $c\in(0,1)$.  The polynomials $M_r(\,\cdot\,;n,c)$, $r\ge0$, are orthogonal for the positive discrete weight
\begin{equation}\label{eq:meixner-weight}
 w_c(m)=\frac{(n)_m}{m!}\,c^m,
 \qquad m=0,1,2,\dots,
\end{equation}
and $M_\ell(\,\cdot\,;n,c)$ has $\ell$ simple zeros
\begin{equation}\label{eq:zeros}
 0<\zeta_1(c)<\cdots<\zeta_\ell(c).
\end{equation}
\end{lemma}

\begin{proof}
We derive the orthogonality from the generating function, since the
derivation also fixes the weight that Section~\ref{sec:zero-motion}
differentiates. The generating function in
\citet[\S9.10]{koekoek2010hypergeometric} is
\begin{equation}\label{eq:meixner-gf}
 \sum_{r=0}^{\infty}\frac{(n)_r}{r!}M_r(m;n,c)\,t^r
 =\left(1-\frac tc\right)^{\!m}(1-t)^{-m-n}.
\end{equation}
To compare different degrees, multiply two copies
of~\eqref{eq:meixner-gf}, at arguments $t$ and $s$ with $|t|,|s|$
small, and sum against $w_c$. Including the factor $c^m$ from
$w_c(m)$, the dependence on $m$ is $(n)_m z^m/m!$, where
\[
 z=\frac{(c-t)(c-s)}{c(1-t)(1-s)},
 \qquad
 1-z=\frac{(1-c)(c-ts)}{c(1-t)(1-s)},
\]
so $\sum_{m\ge0}\frac{(n)_m}{m!}z^m=(1-z)^{-n}$ yields
\begin{equation}\label{eq:gf-diagonal}
 \sum_{m=0}^{\infty}w_c(m)
 \Biggl[\,\sum_{r\ge0}\frac{(n)_r}{r!}M_r(m;n,c)\,t^r\Biggr]
 \Biggl[\,\sum_{q\ge0}\frac{(n)_q}{q!}M_q(m;n,c)\,s^q\Biggr]
 =(1-c)^{-n}\left(1-\frac{ts}c\right)^{\!-n}.
\end{equation}
The right side contains only equal powers $t^rs^r$, so comparing coefficients gives
\begin{equation}\label{eq:orth}
 \sum_{m=0}^{\infty}w_c(m)\,M_r(m;n,c)\,M_q(m;n,c)=0
 \qquad(r\ne q).
\end{equation}

Since $\deg M_r=r$, the polynomial $M_\ell$ is orthogonal to every
polynomial of degree $<\ell$, and the usual sign-change argument
yields~\eqref{eq:zeros}. Suppose $M_\ell$ changed sign at fewer than
$\ell$ points of $(0,\infty)$. Multiplying it by the polynomial of
degree $<\ell$ that vanishes at those points would give a product of
one fixed sign on the support $\{0,1,2,\dots\}$, not identically zero
there, contradicting~\eqref{eq:orth}. Hence $M_\ell$ has $\ell$ sign
changes on $(0,\infty)$; these sign changes account for all of its zeros, which are
therefore simple and positive.
\end{proof}

\subsection{Monotonicity of the Meixner zeros}\label{sec:zero-motion}

By Markov's monotonicity theorem, the zeros~\eqref{eq:zeros} move
monotonically with $c$; \citet[Example~4.5]{castillo2017markov} treat
the Meixner case explicitly. We include the quadrature argument because
it gives the positive formula~\eqref{eq:zeta-deriv} for $\zeta_j'$,
which Section~\ref{sec:factorization} turns
into~\eqref{eq:logpsi-deriv}.

Put $\langle f,g\rangle_c=\sum_{m\ge0}f(m)\,g(m)\,w_c(m)$, let
$p_\ell(u,c)$ be the monic normalization of $M_\ell(u;n,c)$, and write
$p_\ell'=\partial_up_\ell$.

\begin{lemma}[Motion of the zeros]\label{lem:zero-motion}
Each zero $c\mapsto\zeta_j(c)$ is strictly increasing on $(0,1)$.
\end{lemma}

\begin{proof}
Fix $j$ and write $\zeta=\zeta_j(c)$. The coefficients of $p_\ell$ are
rational in $c$, and the zeros are simple by
Lemma~\ref{lem:meixner-orth}, so the implicit function theorem applied
to $p_\ell(\zeta_j(c),c)=0$ shows that $\zeta_j$ is differentiable with
\begin{equation}\label{eq:implicit}
 \partial_cp_\ell(\zeta,c)=-p_\ell'(\zeta,c)\,\zeta_j'(c).
\end{equation}
To find the sign of $\zeta_j'(c)$, we evaluate one inner product in two
ways: differentiating orthogonality in $c$ produces a sum of squares,
and Gaussian quadrature at the zeros isolates the single zero
$\zeta_j$. Put
\begin{equation}\label{eq:q-def}
 q(u,c)=\frac{p_\ell(u,c)}{u-\zeta},
\end{equation}
a monic polynomial of degree $\ell-1$, so that $\langle p_\ell,q\rangle_c=0$.

First, differentiate this identity in $c$; the geometric decay of
$w_c$, locally uniformly in $c$, justifies differentiating term by
term. Since $\partial_c\log w_c(m)=m/c$, and since
$\langle p_\ell,\partial_cq\rangle_c=0$ because $\deg\partial_cq<\ell$,
\begin{equation}\label{eq:diff-orth}
 \langle\partial_cp_\ell,\,q\rangle_c
 =-\Bigl\langle p_\ell\,q,\ \frac mc\Bigr\rangle_c.
\end{equation}
Orthogonality also allows us to subtract $\zeta/c$ in the second argument
on the right of~\eqref{eq:diff-orth}. Substituting
$p_\ell=(m-\zeta)q$ then gives
\begin{equation}\label{eq:neg-sum}
 \langle\partial_cp_\ell,\,q\rangle_c
 =-\frac1c\sum_{m=0}^{\infty}(m-\zeta)^2\,q(m,c)^2\,w_c(m).
\end{equation}

Second, we evaluate the same inner product by Gaussian quadrature at
the nodes~\eqref{eq:zeros}. Let $L_k$ be the Lagrange cardinal
polynomials of the nodes and put $\lambda_k=\langle L_k,1\rangle_c$.
Division by $p_\ell$ shows that $\langle f,1\rangle_c=\sum_k\lambda_kf(\zeta_k)$
whenever $\deg f\le2\ell-1$. Because $p_\ell$
is monic, $\deg\bigl((\partial_cp_\ell)\,q\bigr)\le2\ell-2$. Moreover,
$q(\zeta_k,c)=0$ for $k\ne j$ and $q(\zeta,c)=p_\ell'(\zeta,c)$; hence
\begin{equation}\label{eq:quadrature}
 \langle\partial_cp_\ell,\,q\rangle_c
 =\lambda_j\,\partial_cp_\ell(\zeta,c)\,p_\ell'(\zeta,c).
\end{equation}
Since $\deg L_j^2=2\ell-2$, quadrature gives
$\lambda_j=\langle L_j,L_j\rangle_c>0$. Combining \eqref{eq:implicit},
\eqref{eq:neg-sum}, and \eqref{eq:quadrature},
\begin{equation}\label{eq:zeta-deriv}
 \boxed{\ \zeta_j'(c)
 =\frac{\displaystyle\sum_{m=0}^{\infty}(m-\zeta_j)^2\,q(m,c)^2\,w_c(m)}
       {\displaystyle c\,\lambda_j\,p_\ell'(\zeta_j,c)^2}
 \;>\;0.\ }
\end{equation}
The numerator is positive because the support is infinite while $q(\,\cdot\,,c)$ has only $\ell-1$ zeros.
\end{proof}

\subsection{Factorization of the ratio}\label{sec:factorization}

Factoring $M_\ell$ over its zeros places the multiplicity $i$ in each
factor separately, as $i+\zeta_j(c)$. The Tur\'an-ratio then becomes a
product over the zeros of factors that depend on one zero each, and
Lemma~\ref{lem:zero-motion} decides how each factor moves with $x$.

\begin{proof}[\textbf{Second proof of Corollary~\ref{thm:haddou}}]
Fix $x>1$ and $c=1/x$.  Since $M_\ell(0;n,c)=1$, factoring over the zeros \eqref{eq:zeros} gives
\begin{equation}\label{eq:factorization}
 M_\ell(-i;n,c)=\prod_{j=1}^{\ell}\frac{i+\zeta_j(c)}{\zeta_j(c)}.
\end{equation}
Inserting~\eqref{eq:H-meixner} into~\eqref{eq:Hi-def} cancels the
binomial prefactors and the factors $\zeta_j(c)$, leaving
\begin{equation}\label{eq:psi-product}
 \Psi_{i,n}(x)=\prod_{j=1}^{\ell}
 \frac{\bigl(i+\zeta_j(c)\bigr)^2}
      {\bigl(i-1+\zeta_j(c)\bigr)\bigl(i+1+\zeta_j(c)\bigr)},
 \qquad c=\frac1x.
\end{equation}
Each factor of~\eqref{eq:psi-product} is $e^{\phi_i(\zeta_j(c))}$,
where for $i\ge1$ and $z>0$
\[
 \phi_i(z)=2\log(i+z)-\log(i-1+z)-\log(i+1+z).
\]
With $a=i+z>1$,
\begin{equation}\label{eq:phi-deriv}
 \phi_i'(z)=\frac2a-\frac1{a-1}-\frac1{a+1}
 =-\frac2{a(a^2-1)}<0.
\end{equation}
Since $\log\Psi_{i,n}(x)=\sum_j\phi_i(\zeta_j(1/x))$ and
$\frac{d}{dx}\zeta_j(1/x)=-x^{-2}\zeta_j'(1/x)$,
formulas~\eqref{eq:zeta-deriv} and~\eqref{eq:phi-deriv} give
\begin{equation}\label{eq:logpsi-deriv}
 \boxed{\ \frac{d}{dx}\log\Psi_{i,n}(x)
 =\frac2{x^2}\sum_{j=1}^{\ell}
 \frac{\zeta_j'(1/x)}
      {\bigl(i+\zeta_j(1/x)\bigr)
       \Bigl(\bigl(i+\zeta_j(1/x)\bigr)^2-1\Bigr)}
 \;>\;0.\ }\qedhere
\end{equation}
\end{proof}

\begin{rmk}[Why one row]\label{rmk:one-row}
The proof uses three properties of $i\mapsto H_i(x)$ at fixed $\ell$,
$n$, and $x$: it is a polynomial in $i$, it has only real zeros, all
negative, and these zeros move monotonically in $x$. The first property
holds for every shape. The branching rule gives
\[
 S_{\lambda,i}(x)
 =\sum_{\mu\subseteq\lambda}x^{|\mu|}\,
   s_\mu(1^{[i]})\,s_{\lambda/\mu}(1^{[n-i]}),
\]
and $s_\mu(1^{[m]})=\prod_{(a,b)\in\mu}(m+b-a)/h(a,b)$ is a
polynomial in $m$ \citep[Chapter~I, \S3, Example~4]{macdonald1995},
with $h(a,b)$ the hook length of the cell in row $a$ and column $b$.
Hence $S_{\lambda,i}(x)$ is a polynomial in $i$ of degree $|\lambda|$. The
second property fails already for one column: for $\lambda=(1,1)$ and
$n=3$, the values $3,5,8,12$ at $i=0,1,2,3$ give
$S_{\lambda,i}(2)=\tfrac12(i^2+3i+6)$, with complex zeros. The reason
is a mismatch of parameter ranges. By the generating function
$\sum_k e_k(1^{[n-i]},x^{[i]})z^k=(1+z)^{n-i}(1+xz)^i$, a column is a
Krawtchouk polynomial in $i$,
$e_k(1^{[n-i]},x^{[i]})=\binom nk K_k(i;p,n)$ with $p=1/(1-x)$
\citep[\S9.11]{koekoek2010hypergeometric}, and $p$ lies in the
orthogonality range $0<p<1$ only when $x<0$. For a row, the Meixner
range $0<c<1$ is exactly $x>1$. The factorization over zeros, and with
it the explicit formula~\eqref{eq:logpsi-deriv}, is therefore a one-row
phenomenon; for other shapes Theorem~\ref{thm:schur-turan} rests on
the Toda identity alone.
\end{rmk}

\section{A Tur\'an-ratio conjecture for Macdonald polynomials}\label{sec:macdonald-conjecture}

The identity $P_\lambda(z;q,q)=s_\lambda(z)$ suggests extending Theorem~\ref{thm:schur-turan} to
Macdonald polynomials. At repeated arguments, however, the extension fails already for a single row
(calculations omitted for brevity). A principal specialization gives a different comparison
with an exact connection to elementary symmetric functions.

Fix $0<q,t<1$ and $n\ge2$, and use the $P$-normalization of
\citet[Chapter~VI]{macdonald1995}. Starting from
$\mathbf t_n=(1,t,\ldots,t^{n-1})$, dilate its first $j$ entries:
\begin{equation}\label{eq:macdonald-aligned}
 \mathsf A_j(x)
  =(x,xt,\ldots,xt^{j-1},t^j,\ldots,t^{n-1}),
 \qquad 0\le j\le n,
\end{equation}
where empty blocks are omitted. For a partition $\lambda$ with at most
$n$ parts, put
\begin{equation}\label{eq:macdonald-principal-ratio}
 \widehat B_j(x)
   =\frac{P_\lambda(\mathsf A_j(x);q,t)}
          {P_\lambda(\mathbf t_n;q,t)},\qquad
 \widehat T_i(x)
   =\frac{\widehat B_i(x)^2}
          {\widehat B_{i-1}(x)\widehat B_{i+1}(x)}.
\end{equation}
Then $\widehat B_j(1)=1$, $\widehat B_0(x)=1$, and
$\widehat B_n(x)=x^{|\lambda|}$. The common normalization cancels from
$\widehat T_i$, and $\widehat T_i\to\Tc_{\lambda,i}$ as
$q=t\uparrow1$.

\begin{conj}[Macdonald principal-specialization comparison]\label{conj:macdonald-turan}
Let $0<q,t<1$, $n\ge2$, and pad $\lambda$ with zeros to length $n$.
If $\lambda_1>\lambda_n$, then
\[
 \widehat T_i'(x)>0\qquad(x>1,\ 1\le i<n).
\]
If $\lambda=(r^n)$, with $r\ge0$, all these ratios equal one.
\end{conj}

\begin{rmk}[Complementation]\label{rmk:macdonald-complement}
For $M\ge\lambda_1$, let $\lambda^c=(M-\lambda_n,\dots,M-\lambda_1)$.
Then $P_{\lambda^c}(z)=(z_1\cdots z_n)^M
P_\lambda(z_1^{-1},\dots,z_n^{-1})$
\citep[Chapter~VI, (4.14)]{macdonald1995}, and the inverted alphabet
$\mathsf A_j(x)^{-1}$ is, after reversal, $x^{-1}t^{1-n}\mathsf A_{n-j}(x)$.
By symmetry and homogeneity,
\[
 P_{\lambda^c}(\mathsf A_j(x);q,t)
  =x^{jM-|\lambda|}\,t^{M\binom n2-(n-1)|\lambda|}\,
   P_\lambda(\mathsf A_{n-j}(x);q,t),
\]
and the prefactor, whose exponent is affine in $j$, cancels from the
second difference. Hence $\widehat T_{\lambda^c,i}=\widehat T_{\lambda,n-i}$
for $1\le i<n$. Conjecture~\ref{conj:macdonald-turan} therefore needs
proof for only one shape of each complementary pair, and the rectangles
$(r^n)$ are the fixed points. The same argument gives
$\Tc_{\lambda^c,i}=\Tc_{\lambda,n-i}$.
\end{rmk}

Two results support the conjecture: $\widehat T_i(q^{-1})>1$ when
$\lambda_1>\lambda_n$ (Theorem~\ref{thm:macdonald-first-step}), and
monotonicity near $x=1$, for $n=2$, and at infinity for single rows
(Proposition~\ref{prop:macdonald-partial}).

\subsection{Evaluation duality and the first $q$-step}\label{sec:macdonald-duality}

The alignment~\eqref{eq:macdonald-aligned} lets Macdonald's
evaluation duality turn $\widehat B_j(q^{-1})$ into normalized
elementary symmetric functions, to which the $t$-Newton inequality
applies.

\begin{theorem}[Comparison at the first $q$-step]\label{thm:macdonald-first-step}
Under the hypotheses of Conjecture~\ref{conj:macdonald-turan},
\[
 \widehat T_i(q^{-1})\ge1\qquad(1\le i<n).
\]
Every inequality is strict unless $\lambda=(r^n)$.
\end{theorem}

Pad $\lambda$ to length $n$ and set
$y_\lambda=(q^{-\lambda_1},q^{-\lambda_2}t,\ldots,
q^{-\lambda_n}t^{n-1})$. Macdonald's evaluation symmetry
\citep[Chapter~VI, (6.6)]{macdonald1995} gives, for every integer $k\ge0$,
\begin{equation}\label{eq:macdonald-rectangular-duality}
 \widehat B_j(q^{-k})
   =\frac{P_{(k^j)}(y_\lambda;q,t)}
          {P_{(k^j)}(\mathbf t_n;q,t)},\qquad 0\le j\le n.
\end{equation}
To check the parameters, apply that symmetry at $(q^{-1},t^{-1})$.
The identity $P_\rho(z;q^{-1},t^{-1})=P_\rho(z;q,t)$
\citep[Chapter~VI, (4.14)(iv)]{macdonald1995} restores the original
parameters, and multiplying each evaluation alphabet by $t^{n-1}$
changes its $r$th entry from $q^{-\mu_r}t^{-(n-r)}$ to
$q^{-\mu_r}t^{r-1}$. Homogeneity cancels this scaling in each normalized
evaluation. Taking $\mu=(k^j)$ proves
\eqref{eq:macdonald-rectangular-duality}.

\begin{proof}[Proof of Theorem~\ref{thm:macdonald-first-step}]
At $k=1$, the identity $P_{(1^j)}=e_j$ reduces
\eqref{eq:macdonald-rectangular-duality} to
\begin{equation}\label{eq:macdonald-elementary-duality}
 \widehat B_j(q^{-1})=\frac{e_j(y_\lambda)}{c_j},
 \qquad
 c_j=e_j(\mathbf t_n)
    =t^{j(j-1)/2}\prod_{r=1}^j\frac{1-t^{n-j+r}}{1-t^r}.
\end{equation}
For positive $z_1,\ldots,z_n$ satisfying $z_{r+1}\le t z_r$,
the $t$-Newton inequality states that
\begin{equation}\label{eq:separated-newton}
 \left(\frac{e_i(z)}{c_i}\right)^2
 \ge\frac{e_{i-1}(z)}{c_{i-1}}\frac{e_{i+1}(z)}{c_{i+1}}
 \qquad(1\le i<n).
\end{equation}
Indeed, the polynomial $\prod_r(1+z_r u)$ has $t$-separated negative
zeros, so \citet[Theorem~27]{lamprecht2012suffridge} gives
\eqref{eq:separated-newton}. If $z$ is not a scalar multiple of
$\mathbf t_n$, the polynomial is nonextremal; Theorem~4(c),(d) of the
same paper then makes every interior inequality strict.
Now
\[
 \frac{(y_\lambda)_{r+1}}{(y_\lambda)_r}
   =tq^{\lambda_r-\lambda_{r+1}}\le t,
\]
with equality for every $r$ if and only if $\lambda_1=\lambda_n$.
Substituting $z=y_\lambda$ into~\eqref{eq:separated-newton} proves
the assertion. In the exceptional case,
$P_{(r^n)}(z)=(z_1\cdots z_n)^r$, so
$\widehat B_j(x)=x^{rj}$ and $\widehat T_i(x)=1$.
\end{proof}

Since $\widehat T_i(1)=1$, Conjecture~\ref{conj:macdonald-turan}
predicts $\widehat T_i(q^{-k})>1$ for every $k\ge1$ when
$\lambda_1>\lambda_n$. By~\eqref{eq:macdonald-rectangular-duality},
this says that
$j\mapsto P_{(k^j)}(y_\lambda;q,t)/P_{(k^j)}(\mathbf t_n;q,t)$ is
strictly log-concave on $0\le j\le n$.
Theorem~\ref{thm:macdonald-first-step} is the case $k=1$; for $k\ge2$
we have no proof.

\subsection{Monotonicity near $x=1$, in two variables, and at infinity}\label{sec:macdonald-partial}

\begin{prop}[Partial monotonicity]\label{prop:macdonald-partial}
Let $\lambda$ be padded to length $n$, with $\lambda_1>\lambda_n$.
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=*]
\item $(\log\widehat T_i)'(1)>0$ for $1\le i<n$.
\item If $n=2$, then $\widehat T_1$ is strictly increasing on $(1,\infty)$.
\item If $\lambda=(m)$, then for $2\le i<n$, with $\gamma_i>0$ given
by~\eqref{eq:macdonald-positive-tail},
\[
 x\frac{d}{dx}\log\widehat T_i(x)
   =\frac{\gamma_i}{x}+O(x^{-2})\qquad(x\to\infty),
\]
and for $i=1$ the left side tends to $m$. Hence every $\widehat T_i$ is
strictly increasing for all sufficiently large $x$.
\end{enumerate}
\end{prop}

\begin{proof}
(i) Write $D_r=z_r\partial_{z_r}$. Differentiating the first
$j$ coordinates in~\eqref{eq:macdonald-aligned} and taking the adjacent
second difference gives
\begin{equation}\label{eq:macdonald-initial-slope}
 \left.\frac{d}{dx}\log\widehat T_i(x)\right|_{x=1}
   =(D_i-D_{i+1})\log P_\lambda(\mathbf t_n;q,t)>0.
\end{equation}
To see the sign, the
branching formula
\citep[Chapter~VI, (6.24) and (7.14$'$)]{macdonald1995} gives nonnegative
monomial coefficients. Pair monomials exchanged by coordinates
$i,i+1$. For exponents $a>b$, their contribution to
$(D_i-D_{i+1})P_\lambda(z)$ is a nonnegative multiple of
\[
 (a-b)(z_i^a z_{i+1}^b-z_i^b z_{i+1}^a),
\]
which is positive when $z_i>z_{i+1}$. Since $\lambda_1>\lambda_n$, the
monomial orbit of $\lambda$ contains such a pair with positive
coefficient, and $t^{i-1}>t^i$ gives strictness.

(ii) Put $d=|\lambda|$ and $f(u)=P_\lambda(u,1;q,t)$.
The function $F(v)=e^{-dv/2}f(e^v)$ is a nonnegative combination of
hyperbolic cosines, with a nonconstant term since $\lambda_1>\lambda_2$.
Homogeneity gives
\[
 \widehat T_1(x)
   =\frac{t^d}{P_\lambda(1,t;q,t)^2}
      F(\log(x/t))^2.
\]
For $x>1$ we have $\log(x/t)>0$, so $\widehat T_1$ increases strictly.

(iii) For a single row, the $Q$-normalization makes the generating
function particularly simple. Write
$C_j^{(m)}(x)=Q_{(m)}(\mathsf A_j(x);q,t)$
and $(a;q)_k=\prod_{r=0}^{k-1}(1-aq^r)$, with the analogous infinite
product for $k=\infty$. Macdonald's one-row generating product
\citep[Chapter~VI, (2.8)]{macdonald1995} telescopes along the two aligned
blocks:
\begin{equation}\label{eq:macdonald-row-generating}
 \sum_{m\ge0}C_j^{(m)}(x)z^m
   =\frac{(xt^jz;q)_\infty}{(xz;q)_\infty}
      \frac{(t^nz;q)_\infty}{(t^jz;q)_\infty}.
\end{equation}
The $q$-binomial theorem therefore gives
\begin{equation}\label{eq:macdonald-row-coefficients}
 C_j^{(m)}(x)
  =\sum_{k=0}^m
    \frac{(t^j;q)_k}{(q;q)_k}
    \frac{(t^{n-j};q)_{m-k}}{(q;q)_{m-k}}
    t^{j(m-k)}x^k.
\end{equation}
For fixed $m\ge1$, abbreviate $C_j=C_j^{(m)}$. The scalar relating
$P_{(m)}$ and $Q_{(m)}$ cancels, so
$\widehat T_i=C_i^2/(C_{i-1}C_{i+1})$.

For $j\ge1$, the top two coefficients in
\eqref{eq:macdonald-row-coefficients} give
\begin{equation}\label{eq:macdonald-row-tail}
 C_j(x)=\frac{(t^j;q)_m}{(q;q)_m}x^m
       \left(1+\frac{r_j}{x}+O(x^{-2})\right),\qquad
 r_j=\frac{1-q^m}{1-q}\frac{t^j-t^n}{1-q^{m-1}t^j}.
\end{equation}
Consequently, for $2\le i<n$,
\[
 x\frac{d}{dx}\log\widehat T_i(x)
   =\frac{\gamma_i}{x}+O(x^{-2}),\qquad
 \gamma_i=r_{i-1}+r_{i+1}-2r_i.
\]
Writing $a=t^i$, $b=t^n$, and $c=q^{m-1}$, subtraction yields
\begin{equation}\label{eq:macdonald-positive-tail}
 \gamma_i=\frac{1-q^m}{1-q}
   \frac{a(1-t)^2(1+ac)(1-bc)}
        {(t-ac)(1-ac)(1-act)}>0.
\end{equation}
Every denominator is positive because $ac\le t^i<t$.
For $i=1$, $C_0$ is constant and
$x(d/dx)\log\widehat T_1(x)\to m>0$ instead.
\end{proof}

We note in passing that Conjecture~\ref{conj:macdonald-turan} also
holds for $\lambda=(2)$ and every $n\ge2$, hence by
Remark~\ref{rmk:macdonald-complement} for $\lambda=(2^{n-1})$: there
$P_{(2)}(\mathsf A_j(x);q,t)$ is an explicit quadratic in $x$, and
monotonicity follows from the positivity of a polynomial of degree four
in $x-1$, whose verification we omit.

\noindent\rule{\linewidth}{0.3pt}

\vskip8pt

\noindent\textbf{AI Usage Statement}. The results in this paper were developed through extensive use of
an AI assistant (OpenAI's GPT~Pro). None of the claims were obtained zero-shot: they emerged through
sustained back-and-forth discussion, including failed attempts, counterexamples, audits, and repairs. 

\vskip5pt
\noindent\textbf{Acknowledgements}. The author (SS) thanks the Alexander von Humboldt (AvH) Foundation for their generous support via a Humboldt Professorship in AI.

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\end{document}