An Inequality $\left(\sum\limits_{k=1}^n a_k^{1/2}\right)^2\le\left(\sum\limits_{k=1}^n a_k^{1/3}\right)^3$

Question

Why is $\left(\sum\limits_{k=1}^n a_k^{1/2}\right)^2\le\left(\sum\limits_{k=1}^n a_k^{1/3}\right)^3$ with $a_k$ nonnegative

Writing

$\left(\sum\limits_{k=1}^n a_k^{1/2}\right)^2=\left(\sum\limits_{k=1}^n a_ka_k^{-1/2}\right)^2$

$\left(\sum\limits_{k=1}^n a_k^{1/3}\right)^3=\left(\sum\limits_{k=1}^n a_ka_k^{-2/3}\right)^3$

and assuming $\sum\limits_{k=1}^n a_k=1$ the inequality is equivalent to,

$\left(\sum\limits_{k=1}^n a_ka_k^{-1/2}\right)^{-2}\ge\left(\sum\limits_{k=1}^n a_ka_k^{-2/3}\right)^{-3}$

This is almost the power mean inequality, if the exponent on the RHS were $-\frac 32$ instead of $-3$ but if $\sum\limits_{k=1}^n a_k=1$ then $\sum\limits_{k=1}^n a_k^{1/3}\ge1$ hence $\left(\sum\limits_{k=1}^n a_ka_k^{-2/3}\right)^{-\frac32}\ge \left(\sum\limits_{k=1}^n a_ka_k^{-2/3}\right)^{-3}$ so we're done

Is there a possibility to solve this more directly

Answer 1

We need to prove that $$3\sum\limits_{i\neq j}\sqrt[3]{a_i^2a_j}+6\sum\limits_{i\neq j,j\neq k,k\neq i }\sqrt[3]{a_ia_ja_k}\geq2\sum\limits_{i\neq j}\sqrt{a_ia_j},$$ which is obvious because by AM-GM $\sqrt[3]{a_i^2a_j}+\sqrt[3]{a_j^2a_i}\geq2\sqrt{a_ia_j}$.

Done!

Answer 2

For simplicity, let $n=2$. It is easy to generalize to the case of $n>2$. Let $$ a_1=r^2b_1^4, a_2=r^2b_2^4$$ such that $$ b_1^2+b_2^2=1,b_1,b_2\ge0. $$ Then the inequality becomes $$ b_1^{\frac43}+b_2^{\frac43}\ge 1 $$ which is easy to prove. In fact, noting $0\le b_1,b_2\le1$, one has $$ b_1^{\frac43}+b_2^{\frac43}\ge b_1^2+b_2^2=1. $$