# 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.

$\boxed{4}$ (Lastnightstar): If $x_1,x_2,\ldots,x_n\in\mathbb{R}^{+}$ such that $\displaystyle \frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}=n$ then

$x_1+x_2+\cdots+x_n\leq nx_1x_2{\cdots}x_n$

Proof: We use the same notations as used in $\boxed{3}$. The condition is equivalent to $s_{n-1}=s_n$
Hence the inequality after homogenization becomes

$s_1 \le s_n\left(\frac{s_{n-1}}{s_n}\right)^{n-1} \iff s_{n-1}^{n-1} \ge s_n^{n-2}s_1$

which is $B_{(n-1,n-1)}$

$\boxed{5}$ (Lastnightstar): If $x_1,x_2,\ldots,x_n\in\mathbb{R}^+$ then

$\left(\sum\limits_{i=1}^n x_i \right)^{n+2}\ge n^{n+1}\prod\limits_{i=1}^n x_i\sum\limits_{i=1}^n x_i^2$

Proof: Using the same notations by AM-GM

$\left(\sum\limits_{i=1}^n x_i\right)^2=\sum\limits_{i=1}^n x_i^2+ n(n-1)s_2 \ge n\sqrt[n]{\left(ns_2\right)^{n-1}\sum\limits_{i=1}^n x_i^2}$

Thus

$\displaystyle n^{2n}s_1^{2n} \ge n^n\cdot n^{n-1}s_2^{n-1}\sum_{i=1}^n x_i^2$

Again by $A_{(n-1, 2)}$ we have $s_2^{n-1} \ge s_1^{n-2}s_n$. Using this we have

$\displaystyle n^{2n}s_1^{2n} \ge n^n\cdot n^{n-1}s_2^{n-1}\sum_{i=1}^n x_i^2 \ge n^{n-1}s_1^{n-2}s_n\sum_{i=1}^n x_i^2 \implies ns_1^{n+2} \ge s_n\sum_{i=1}^n x_i^2$

which is equivalent to

$\displaystyle (x_1+x_2+\cdots+x_n)^{n+2} \ge n^{n+1}\prod_{i=1}^n x_i \sum_{i=1}^n x_i^2$

which is precisely what we want to show. Hence the proof is complete. $\square$

Note that $\boxed{5}$ for three variables becomes

$(x_1+x_2+x_3)^5 \ge 81x_1x_2x_3(x_1^2+x_2^2+x_3^2)$

which is the well-known Vasc inequality.

Let us use the notations used by Lastnightstar from here.

Let $x_1,x_2,\ldots, x_n$ be nonnegative reals. Let

$\displaystyle \nu_{(m,r)}=\frac{1}{\binom{n}{m}}\sum_{1\leq{ i_1}<{i_2}<{\cdots}<{i_{m}}\leq{ n}}{(x_{i_1}x_{i_2}{\cdots}x_{i_m})^r}$

denote the general arithmetic mean of the symmetric sum of variables.

Then maclaurin’s inequality is equivalent to $\displaystyle \nu_{(1,1)}\ge \sqrt{\nu_{(2,1)}} \ge \sqrt[3]{\nu_{(3,1)}} \ge \cdots \ge \sqrt[n]{\nu_{(n,1)}}$ and Newton’s inequality is equivalent to $\displaystyle \nu_{(k,1)}^2 \ge \nu_{(k+1,1)}\nu_{(k-1,1)}$.

Under this notations the two inequalities in $\boxed{3}$ become $\displaystyle \nu_{(m,1)}^k \ge \nu_{(m-1,1)}^{k-1}\nu_{(m+k-1,1)}$ and $\displaystyle \nu_{(m,1)}^k \ge \nu_{(m+1,1)}^{k-1}\nu_{(m-k+1,1)}$ respectively.

$\boxed{6}$: Let $\displaystyle x_1,x_2,\ldots, x_n\in\mathbb{R}^+, n\geq m+k-1\geq m\geq 1$ and $\displaystyle n,m,k\in\mathbb{N}$, then show that

$\displaystyle\nu_{(1,m-1)}^{k-1}\nu_{(1,m+k-1)}-\nu_{(1,m)}^k\geq 0$

Proof: The inequality is equivalent to

$\displaystyle\left(\sum_{i=1}^n x_i^{m-1}\right)^{k-1}\left(\sum_{i=1}^n x_i^{m+k-1}\right) \geq \left(\sum_{i=1}^n x_i^m\right)^k$

Which is true because by Holder:

\displaystyle\begin{aligned}\left(\sum_{i=1}^n x_i^{m-1}\right)^{k-1} \left(\sum_{i=1}^n x_i^{m+k-1}\right) & \ge \left(\sum_{i=1}^n x_i^{\frac{(m-1)(k-1)+m+k-1}{k}}\right)^k \\ & = \left(\sum_{i=1}^n x_i^m\right)^k \end{aligned}

Similarly we can prove:

$\boxed{7}$: If $x_1,x_2,\ldots, x_n\in\mathbb{R}^+, n\geq m\geq m-k+1\geq 0; n,m,k\in\mathbb{N}$, then

$\displaystyle \nu_{(1,m+1)}^{k-1}\nu_{(1,m-k+1)}-\nu_{(1,m)}^k\geq 0$

$\boxed{8}$ Show that $\displaystyle \nu_{(1,p-1)}\nu_{(1,q+1)}+\nu_{(1,p+1)}\nu_{(1,q-1)}\geq 2\nu_{(1,p)}\nu_{(1,q)}$ for suitable values of $p,q$.

Proof:  By AM-GM and Cauchy Schwarz we have

\displaystyle\begin{aligned}\nu_{(1,p-1)}\nu_{(1,q+1)}+\nu_{(1,p+1)}\nu_{(1,q-1)} & \ge 2\sqrt{\nu_{(1,p-1)}\nu_{(1,q+1)}\nu_{(1,p+1)}\nu_{(1,q-1)}} \\ & \ge 2\sqrt{\nu_{(1,p)}^2\nu_{(1,q)}^2} \\ & = 2\nu_{(1,p)}\nu_{(1,q)} \end{aligned}

$\boxed{9}$ (Lastnightstar): Let $f_{(3,1)}=\nu_{(1,3)}-\nu_{(1,2)}\nu_{(1,1)}, \ \ \ f_{(3,2)}=\nu_{(2,1)}\nu_{(1,1)}-\nu_{(3,1)}$ and let $g_{(3,1)}=2f_{(3,1)}-(n-1)f_{(3,2)}$. Show that $g_{(3,1)} \ge 0$

Proof: Let us prove for $3$ variables. For $3$ variables the inequality is equivalent to

$\displaystyle 2\left(\frac13\sum_{cyc} x^3 -\frac19\sum_{cyc} x^2\sum_{cyc} x\right) \ge 2\left(\frac19\sum_{cyc} xy\sum_{cyc} x -xyz\right)$

$\displaystyle\iff 2(2\sum_{cyc} x^3 -\sum_{cyc} xy(x+y)) \ge 2(\sum_{cyc} xy(x+y) -6xyz)$

$\displaystyle\iff \sum_{cyc} x^3 +3xyz \ge \sum_{cyc} xy(x+y)$

which Schur’s inequality of $3$ degree.

For $n > 3$ variables,

$\displaystyle 2\left(\frac1n\sum_{i=1}^n x_i^3 - \frac1{n^2}\sum_{i=1}^n x_i^2\sum_{i=1}^n x_i\right)$

$\displaystyle \ge (n-1)\left(\frac{1}{n}\sum_{i=1}^n x_i\frac{1}{\binom{n}{2}}\sum_{1\le i

$\displaystyle\iff 2\left(n\sum_{i=1}^n x_i^3-\sum_{i=1}^n x_i^2\sum_{i=1}^n x_i\right) \ge \left(2\sum_{i=1}^n x_i\sum_{1\le i

$\displaystyle (n-2)\left(n\sum_{i=1}^n x_i^3-\sum_{i=1}^n x_i^2\sum_{i=1}^n x_i\right) \ge \left((n-2)\sum_{i=1}^n x_1\sum_{1\le i

Let us simplify the left side:

$\displaystyle (n-2)\left(n\sum_{i=1}^n x_i^3 -\sum_{i=1}^n x_i^2\sum_{i=1}^n x_i\right)= \sum_{i=1}^n x_i^2(n-2)(nx_i-\sum_{i=1}^n x_i)$

Note that

\displaystyle \begin{aligned}\underset{i\neq k, j \neq k}{\sum_{1\le i

Hence

\displaystyle\begin{aligned}(n-2)\left(n\sum_{i=1}^n x_i^3 -\sum_{i=1}^n x_i^2\sum_{i=1}^n x_1\right) & = \sum_{k=1}^n x_k^2\left(\underset{i\neq k, j \neq k}{\sum_{1\le i

where

$\displaystyle A_{(i.j,k)}=x_i^2(2x_i-x_j-x_k)+x_j^2(2x_j-x_k-x_i)+x_k^2(2x_k-x_i-x_j)$

Now let us simplify the right side:

$\displaystyle (n-2)\sum_{k=1}^n x_k\sum_{1\le i

Note that

\displaystyle\begin{aligned}& \sum_{k=1}^n x_k\sum_{1\le i

Therefore

\displaystyle\begin{aligned} & (n-2)\sum_{k=1}^n x_k\sum_{1\le i

Note that $A_{(i,j,k)} \ge B_{(i,j,k)}$ Since it is exactly equivalent to $g_{(3,1)} \ge 0$ for $3$ variables namely $i,j,k$, i.e., Schur’s inequality of $3$ degree.
Hence we have

$\displaystyle g_{(3,1)}=\sum_{1\le i

$\boxed{10}$ Show that

$\displaystyle 3n\cdot\nu_{(1,1)}(\nu_{(2,1)}\nu_{(1,1)}-\nu_{(3,1)})\geq 2(n-1)(\nu_{(3,1)}\nu_{(1,1)}-\nu_{(4,1)})$

Proof: It can be proved similarly like the last one.

To be expanded…..

The full thing is already available in my old blog:

http://www.artofproblemsolving.com/community/c2426h1043311_symmetric_n_variable_inequalities