Let $x_1,\dots, x_n$ be $n$ variables. For brevity I'll use the following notation: for $I \subset \{1,\dots , n\}$ define $$ x_I := \prod_{i \in I} x_i. $$ Let $e_k(x_1,\dots, x_n)$ be an elementary symmetric polynomial. That is $$ e_k(x_1,\dots, x_n) = \sum_{\substack{I \in \binom{[n]}{k}}} x_I. $$ Suppose that we have the following conditions on $x_1,\dots, x_n$: $$ \sum_{i=1}^n x_i = 1 \qquad \text{and} \qquad x_i \geq 0 $$ I have been trying to prove the following inequality without Lagrange multipliers: $$ e_k(x_1,\dots x_n) \leq \frac{\binom{n}{k}}{n^{k}} $$

I have had two difficulties:

(1) I have made an (albeit feeble) attempt using AM-GM. Any hints on how one might proceed without Lagrange Multipliers would be very much appreciated.

(2) Additionally, when one does use Lagrange Multipliers I have been having trouble arguing that the maximum does indeed occur when $$ x_1 = \cdots = x_n = \frac{1}{n}. $$

Thanks in advance for your help.

  • 1
    $\begingroup$ For $k=2$, you have $2e_2=(\sum_k x_k)^2-\sum_k x_k^2=1-\sum_k x_k^2$. Next, Cauchy-Schwarz gives us $||X||^2||Y||^2 \geq <X,Y>^2$ where $X=(x_1,\ldots,x_n)$ and $Y=(1,\ldots,1)$, so $(n^2)\sum_k x_k^2 \geq (\sum_k x_k)^2=1$ with equality iff $X$ and $Y$ are proportional. So, the $k=2$ case is finished. $\endgroup$ Jan 21, 2023 at 8:42
  • 1
    $\begingroup$ @JazzGuitar7 Since variables $x_1,\cdots, x_n$ are basically free, Lagrange multipliers does not help more than taking derivative directly. $\endgroup$
    – Apass.Jack
    Jan 22, 2023 at 3:46

4 Answers 4


You can proceed by induction over $n$. The claim holds for $n=1$. Since

$$ e_k(x_1,\ldots,x_{n+1})=e_k(x_1,\ldots,x_n)+x_{n+1}e_{k-1}(1,\ldots,x_n)\;, $$

assuming that the claim holds for $n$ and maximizing this for fixed $x_{n+1}$ by scaling the constraint in the claim for $n$ by $1-x_{n+1}$ shows that the maximum is

$$ \frac{\binom nk}{n^k}(1-x_{n+1})^k+x_{n+1}\frac{\binom n{k-1}}{n^{k-1}}(1-x_{n+1})^{k-1}\\=\frac{\binom nk}{n^k}(1-x_{n+1})^{k-1}\left(1-x_{n+1}+\frac{kn}{n-k+1}\cdot x_{n+1}\right)\;. $$

Setting the derivative with respect to $x_{n+1}$ to zero yields

$$ -(k-1)\left(1-x_{n+1}+\frac{kn}{n-k+1}\cdot x_{n+1}\right)+(1-x_{n+1})\left(\frac{kn}{n-k+1}-1\right)=0\;. $$

You can check by substitution that the solution is $x_{n+1}=\frac1{n+1}$ as required.

  • 2
    $\begingroup$ Nice. Since what you are maximizing is a product, perhaps there is a nice rewriting of it as another product, that allows to show the result using only AM-GM and not the derivative $\endgroup$ Jan 21, 2023 at 10:48

All $x_i$ are nonnegative. For brevity, I will write $e_k(x_1,\cdots, x_n)$ as $e_{n,k}$ when the variables are $x_1,\cdots, x_n$.

First proof, a refined inequality.

Let $f$ be a sequence of nonincreasing positive integers $(f_1, \cdots, f_\ell)$ such that $\sum_if_i=k$. Consider all distinct terms of the form $x_{i_1}^{f_1}x_{i_2}^{f_2}\cdots x_{i_\ell}^{f_\ell}$ where $\{i_1,i_2,\cdots, i_\ell\}$ is a subset of $[n]$ with $\ell$ elements. The sum of all those terms is called the monomial symmetric polynomial of type $f$ over $x_1, \cdots, x_n$, denoted by $m_{n,f}$.

For example, $m_{4,\{2,1\}}=x_1^2x_2+x_1^2x_3+x_1^2x_4$ $+x_2^2x_1+x_2^2x_3 + x_2^2x_4$ $+x_3^2x_1+x_3^2x_2+ x_3^2x_4$ $+x_4^2x_1+x_4^2x_2+x_4^2x_3$. The basic symmetrical polynomail $e_{n,k}$ is none other than $m_{n,\underbrace{\{1,1,\cdots,1\}}_{k\ 1's}}$, which has $n\choose k$ terms.

A refined inequality: Let $m_{n,f}$ be a monomial symmetric polynomial with $\sum_{f_i\in f}f_i=k\le n$. Then $$\frac{m_{n,f}}{\|m_{n,f}\|}\ge \frac{e_{n,k}}{\|e_{n,k}\|}$$ where $\|p\|$ is the number of terms in a monomial symmetric polynomial $p$. The equality holds if either $m_{n,f}=e_{n,k}$ or all terms in $m_{n,f}$ are equal (which means either all terms are zero or all $x_i$ are equal).

This is the special case of Muirhead inequality with $b=(1,1,\cdots,1,0,0,\cdots,0)$, where there are $k$ $1$'s and $n-k$ $0$'s in $b$.

Explanation of a proof: Let me illustrate the simple idea of the proof in the the special case of $n=4, f=(2,1)$, $\frac{m_{4,\{2,1\}}}{12}\ge \frac{e_{4,3}}4$. By AM-GM, we have the following. $$\frac{x_1^2x_2+x_1^2x_3+x_2^2x_1+x_2^2x_3+x_3^2x_1+x_3^2x_2}6\ge x_1x_2x_3$$ $$\frac{x_1^2x_2+x_1^2x_4+x_2^2x_1+x_2^2x_4+x_4^2x_1+x_4^2x_2}6\ge x_1x_2x_4$$ $$\frac{x_1^2x_3+x_1^2x_4+x_3^2x_1+x_3^2x_4+x_4^2x_1+x_4^2x_3}6\ge x_1x_3x_4$$ $$\frac{x_2^2x_3+x_2^2x_4+x_3^2x_2+x_3^2x_4+x_4^2x_2+x_4^2x_3}6\ge x_2x_3x_4$$ Adding them together, we get the desired specialized inequality.

For general $n$ and $f$, we can apply AM-GM to all terms in $m_{n,f}$ that only involve $x_{i_1},x_{i_2},\cdots,x_{i_k}$, for the term $x_{i_1}x_{i_2}\cdots x_{i_k}$. Adding all inequalities thus obtained for every term in $e_{n,k}$, we will prove the inequality. The actual computation is omitted here.

The refined inequality above says that for any monomial symmetric polynomial over $x_1,\cdots, x_n$ of any type, the average of its terms is no less than the average of terms in $e_{n,k}$. Since (the expansion of) $(\sum_ix_i)^k$ is a positive linear combination of those kinds of polynomials, the average of all $n^k$ terms in it is no less than the average of terms in $e_{n,k}$. That is $$\frac{(\sum_ix_i)^k}{n^k}\ge \frac{e_{n,k}}{n\choose k}.$$ The equality holds if $k=1$ or each $x_i$ is $\frac1n$. If the equality holds, then $\frac{m_{n, \{k\}}}n=e_{n,k}$, which implies $k=1$ or all $x_i$ are equal, i.e., to $\frac1n$.

Second proof, induction on the number of variables

Claim. For nonnegative numbers $x_1, \cdots, x_n$ and integer $1\le k\le n$, we have $$ e_{n,k} \leq \binom{n}{k}\left(\frac{\sum_i x_i}n\right)^{k} $$ where the equality holds iff $k=1$ or all $x_i$'s are equal. The inequality in the question is the case when $\sum_ix_i=1$.

Let us prove the claim by induction on $n$. The base case when $n=1$ is trivial.

Suppose it is true for smaller $n$. Assume $n\ge2$. Assume $k\ge2$; otherwise with $k=1$, the situation is trivial. Let $S=\sum_{i=1}^n x_i$.

$$\begin{aligned} e_{n,k}&=e_{n-1,k-1}x_n+ e_{n-1,k}\\ &\le \binom{n-1}{k-1}\left(\frac{S-x_n}{n-1}\right)^{k-1}x_n + \binom{n-1}{k}\left(\frac{S-x_n}{n-1}\right)^{k}\\ &= \frac1k\binom{n-1}{k-1}\left(\frac1{n-1}\right)^{k}(S-x_n)^{k-1}(k(n-1)x_n+(n-k)(S-x_n))\\ &= \frac1k\binom{n-1}{k-1}\left(\frac1{n-1}\right)^{k}(S-x_n)^{k-1}((k-1)nx_n+(n-k)S)\\ \end{aligned}$$ where the inequality above comes from the induction hypothesis. $$\begin{aligned} (S-x_n)^{k-1}&((k-1)nx_n+(n-k)S)\\ &=\frac1{n^{k-1}}(n(S-x_n))^{k-1}((k-1)nx_n+(n-k)S)^1\\ &\le \frac1{n^{k-1}}\left(\frac{(k-1)\cdot n(S-x_n)+1\cdot((k-1)nx_n+(n-k)S)}{k}\right)^k\\ &= \frac1{n^{k-1}}\left((n-1)S\right)^k \end{aligned}$$ where the inequality is the AM-GM inequality. So, $$e_{n,k}\le\frac1k\binom{n-1}{k-1}\left(\frac1{n-1}\right)^{k}\frac1{n^{k-1}}\left((n-1)S\right)^k={n\choose k}\left(\frac Sn\right)^k$$

The equality in the inequality holds iff $x_1,\cdots, x_{n-1}$ are all equal by induction hypothesis and $n(S-x_n)=(k-1)nx_n+(n-k)S$ from AM-GM, which means $x_n=\frac S{n}$. Hence, we must have $x_i=\frac Sn$ for all $i$. Induction on $n$ is complete.

Third proof, assuming the maximum can be reached.

Suppose $e_k(x_1,\dots x_n)$ reaches the maximum when $(x_1, \cdots, x_n)=(a_1, \cdots, a_n)$. WLOG assume $n\ge3$ and $a_1\le a_2\cdots\le a_n$.

Since $e_k(a_1,\dots, a_n)\ge e_k(\frac1n,\frac1n,\cdots,\frac1n)>0$, we must have $e_{k-2}(a_2, a_3, \cdots, a_{n-1})>0$. .

If $a_1<a_n$, then $$e_k(\frac{a_1+a_n}2, a_2,a_3, \cdots, a_{n-1}, \frac{a_1+a_n}2)-e_k(a_1,\cdots, a_n)\\=e_{k-2}(a_2, a_3, \cdots, a_{n-1})\left(\frac{a_n-a_1}2\right)^2>0,$$ which contradicts with the maximality of $e_k(a_1,\cdots, a_n)$. Hence $a_1=a_n$, i.e., all $a_i$ are equal. The wanted inequality follows easily.

This proof is basically a copy of Ewan Delanoy's answer.


Here is yet another proof. Let $i\lt j$ be any two indices in $[|1..n|]$. Denote by $t_1 \lt t_2 \lt\ldots,t_{n-2}$ the other indices in $[|1..n|]$ and let $y_k=x_{t_k}$.

We can write

$$ e_k(x_1,\ldots,x_n)=e_{k}(y_1,\ldots,y_{n-2})+e_{k-1}(y_1,\ldots,y_{n-2})(x_i+x_j)+ e_{k-2}(y_1,\ldots,y_{n-2})x_ix_j $$

From this, we see that if $(x_i,x_j)$ is replaced by $(\frac{x_i+x_j}{2},\frac{x_i+x_j}{2})$, the sum stays the same but the product gets larger, so we get a larger value. In fact, we get a strictly larger value, unless all the coordinates are zero (indeed, if $e_{k-2}(y_1,\ldots,y_{n-2})=0$, then all the monomials appearing in it must also be zero, hence $e_{k-1}(y_1,\ldots,y_{n-2})=0$ and $e_k(x_1,\ldots,x_n)=0$ also).

In particular, if $(x_1,\ldots,x_n)$ is a point where the maximum is attained, we must have $x_i=x_j$. Since this holds for any two indices, all the coordinates are equal, and the rest is easy.


Your inequality follow immediately from the Maclaurin's inequality.

The Maclaurin inequality we can prove by the Rolle's theorem.


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