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

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!
