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 

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
$$ b_1^{\frac43}+b_2^{\frac43}\ge b_1^2+b_2^2=1. $$
