Fix a natural number $n\geq 1$. Let $a_1, \ldots, a_n$ be $n$ real numbers such that $a_i>0$ for each $i$. Show that for each natural $k$ with $0\leq k\leq n$ $$e_k(a_1,\ldots, a_n)\geq\binom{n}{k}(a_1\cdot\cdots\cdot a_n)^{k/n}$$ where $e_k(x_1, \ldots, x_n)$ denotes the $k$th symmetric polynomial with $n$ arguments and where we make the convention that $e_0(x_1,\ldots, x_n)=1$.

Here's the proof I gave:

It suffices to prove the claim assuming that $a_1\cdot\cdots\cdot a_n=1$. For if this is not the case, then setting $\overline{a_i}=a_i/(a_1\cdot\cdots\cdot a_n)^{1/n}$ will give $$\overline{a_1}\cdot\cdots\cdot\overline{a_n}=1$$ and the general claim will follow from the fact that $$e_k(\overline{a_1},\ldots,\overline{a_n})=\frac{1}{(a_1\cdot\cdots\cdot a_n)^{k/n}} e_k(a_1,\ldots,a_n)$$

Thus we prove that if $a_1\cdot\cdots\cdot a_n=1$ then $e_k(a_1,\ldots,a_n)\geq\binom{n}{k}$. We induct on $n$ (we go through Pascal's Triangle row-by-row left-to-right). Note that this claim is trivial for $k=0$ or $k=n$. Thus when $n=1$, the claim holds for $k=0$ and $k=n$. Now assume the claim is true for the natural number $m$ and all natural $k$ with $0\leq k\leq m$; we show that the claim is true for the natural number $m+1$ and all natural $k$ with $0\leq k\leq m+1$. Since $$e_0(a_1,\ldots, a_{m+1})=1=\binom{m+1}{0}$$ the claim follows trivially. And since $$e_1(a_1,\ldots, a_{m+1})=a_1+\cdots+a_{m+1}\geq m+1$$ is a special case of the Arithmetic-Geometric Mean Inequality we have that this case holds as well. Without loss of generality we can assume that $a_1\leq \cdots\leq a_{m+1}$. This ordering implies that $a_{m+1}\geq 1$ since a product of numbers strictly greater than 0 and strictly less than 1 cannot multiply to 1. Now note that $$e_{k+1}(a_1\ldots, a_{m+1})=e_{k+1}(a_1,\ldots,a_m)+a_{m+1}e_{k}(a_1,\ldots,a_m)$$ for $1\leq k+1\leq m$. Thus $$e_{k+1}(a_1,\ldots, a_{m+1})\geq\binom{m}{k+1}+a_{m+1}\binom{m}{k}\geq\binom{m}{k+1}+\binom{m}{k}=\binom{m+1}{k+1}$$ And since $e_{m+1}(a_1,\ldots,a_{m+1})=a_1\cdot\cdots\cdot a_{m+1}=1=\binom{m+1}{m+1}$, the claim follows by induction.