Proof of an (Elementary) Inequality Involving Elementary Symmetric Polynomials 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.
 A: 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.
A: 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.
A: Your inequality follow immediately from the Maclaurin's  inequality.
The Maclaurin inequality we can prove by the Rolle's theorem.
