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



\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}

Proof: See

\boxed{2} Maclaurin’s Inequality:┬áLet 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<j}{\sum_{i=1}^n\sum_{j=1}^n} x_ix_j

\displaystyle \Longleftrightarrow (n-1)\left[\sum_{i=1}^n x_i^2+2\underset{i<j}{\sum_{i=1}^n\sum_{j=1}^n} x_ix_j\right] \ge 2n\underset{i<j}{\sum_{i=1}^n\sum_{j=1}^n} x_ix_j

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

\displaystyle \Longleftrightarrow \underset{i<j}{\sum_{i=1}^n\sum_{j=1}^n} (x_i-x_j)^2 \geq 0

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}


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)})


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.

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