# Symmetric Inequalities

Symmetric Sums: Let $x_1,x_2, \ldots, x_n$ be $n$ real numbers. We define symmetric sum $s_k$ as the coefficient of $x^{n-k}$ in the polynomial

$P(x)=(x_1+x)(x_2+x)(x_3+x)\cdots (x_n+x)$

and symmetric average is defined by $d_k=\frac{s_k}{\binom{n}k}$. For example for $4$ variables we have

$s_1=x_1+x_2+x_3+x_4$
$s_2=x_1x_2+x_2x_3+x_3x_4+x_4x_1+x_1x_3+x_2x_4$
$s_3=x_2x_3x_4+x_1x_3x_4+x_1x_2x_4+x_1x_2x_3$
$s_4=x_1x_2x_3x_4$

$\boxed{1}$ Newton’s Inequality: Let $x_1,x_2,\cdots,x_n$ be $n$ non-negative reals. Then for all $k \in \{1,2,\cdots,n-1\}$ we have

$d_k^2 \ge d_{k-1}d_{k+1}$

$\boxed{2}$ Maclaurin’s Inequality: Let $x_1,x_2,\ldots,x_n$ be $n$ nonnegative reals. Let their symmetric averages be $d_1,d_2,\ldots,d_n$. Then we have

$d_1 \ge \sqrt{d_2} \ge \sqrt[3]{d_3} \ge \cdots \ge \sqrt[n]{d_n}$

Proof: We will use induction and Newton’s Inequality. Let us prove that $d_1\ge \sqrt{d_2}$ holds. This inequality is equivalent to

$\displaystyle (x_1+x_2+\cdots+x_n)^2 \ge \frac{2n}{n-1}\underset{i

$\displaystyle \Longleftrightarrow (n-1)\left[\sum_{i=1}^n x_i^2+2\underset{i

$\displaystyle \Longleftrightarrow (n-1)\sum_{i=1}^n x_i^2 \ge 2\underset{i

$\displaystyle \Longleftrightarrow \underset{i

which is obvious.

Suppose the chain of inequalities is true upto $\sqrt[k-1]{d_{k-1}} \ge \sqrt[k]{d_k}$. Then by applying Newton’s inequality we have

$d_{k-1}^k \ge d_k^{k-1} \ge \left[\sqrt{d_{k-1}d_{k+1}}\right]^{k-1}\implies d_{k-1}^{2k} \ge d_{k-1}^{k-1}d_{k+1}^{k-1} \implies d_{k-1}^{k+1} \ge d_{k+1}^{k-1}$

Hence

$d_k^{2(k+1)} \ge d_{k-1}^{k+1}d_{k+1}^{k+1} \ge d_{k+1}^{k-1+k+1} \implies \sqrt[k]{d_k} \ge \sqrt[k+1]{d_{k+1}}$

Thus by induction the proof is complete.

$\boxed{3}$ (own): Let $s_k$ denote the arithmetic mean of $k$-th degree symmetric sum of $n$ (positive) variables. If $k \ge 2$ and $m+1 > k$ for $B$ and $m+k-1 \le n$ for $A$ we have the following inequalities:

$s_m^k \ge s_{m-1}^{k-1}s_{m+k-1} \ \ \ \left[A_{(k,m)}\right], \ \ \ s_m^k \ge s_{m+1}^{k-1}s_{m-k+1} \ \ \ \left[B_{(k,m)}\right]$

Proof: To prove this we use induction on $k$.
For $k=2$, both inequalities are equivalent to $s_m^2 \ge s_{m-1}s_{m+1}$ which is Newton’s inequality. Let us suppose both holds for all $k \le t$
Then for $k=t+1$

$s_m^{t} \stackrel{A_{(t,m)}}{\ge} s_{m-1}^{t-1}s_{m+t-1} = s_{m-1}^{t-1}\sqrt[t]{s_{m+t-1}^t} \stackrel{B_{(t,m+t-1)}}{\ge} s_{m-1}^{t-1}\sqrt[t]{s_{m+t}^{t-1}s_m}$

$\implies s_m^{t+1} \ge s_{m-1}^ts_{m+t} \ \ \ (A_{(t+1,m)})$

And

$s_m^t \stackrel{B_{(t,m)}}{\ge} s_{m+1}^{t-1}s_{m-t+1}=s_{m+1}^{t-1}\sqrt[t]{s_{m-t+1}^t} \stackrel{A_{(t,m-t+1)}}{\ge} s_{m+1}^{t-1}\sqrt[t]{s_{m-t}^{t-1}s_{m}}$

$\implies s_m^{t+1} \ge s_{m+1}^ts_{m-t} \ \ \ (B_{(t+1, m)})$

Hence by induction the two inequalities are true.