# How do I prove the inequality $(\sum a^3)^2 \leq (\sum a^2)^3$?

Let $$a_1, \dots, a_n \in \mathbb{R}.$$ I wish to show that $$(\sum_{i=1}^n a_i^3)^2 \leq (\sum_{i=1}^n a_i^2)^3$$ in order to prove another statement. But I cannot see how to prove this, if at all the inequality is true. Can anyone help me?

Re-write the desired inequality as $$\left(\sum^n_{i=1} a_i^{3}\right)^{1/3} \le \left( \sum^n_{i=1} a_i^2\right)^{1/2}.$$ First assume that $$\left( \sum^n_{i=1} a_i^2\right)^{1/2} = 1$$. Then $$\lvert a_i \rvert \le 1$$ for all $$i$$, and thus $$a_i^3 \le a_i^2$$ and so $$\sum^n_{i=1} a_i^3 \le \sum^n_{i=1} a_i^2 = 1 \,\,\,\, \implies \,\,\,\, \left(\sum^n_{i=1} a_i^3\right)^{1/3} \le 1.$$ If $$\left( \sum^n_{i=1} a_i^2\right)^{1/2} \neq 1$$, then define $$A =\left( \sum^n_{i=1} a_i^2\right)^{1/2}$$ and apply this previous result to $$\bar a_i = a_i/A$$. Then you'll find $$\left( \sum^n_{i=1}\frac{a_i^3}{A^3}\right)^{1/3} \le 1 \,\,\,\,\,\, \implies \,\,\,\,\,\, \left( \sum^n_{i=1}a_i^3\right)^{1/3} \le A := \left(\sum^n_{i=1} a_i^2 \right)^{1/2}.$$ (Of course, you cannot divide by $$A$$ if $$A = 0$$, but in that case the whole vector is zero and the inequality is trivial).
Now, let $$a_i^2=x_i.$$
Thus, we need to prove that $$\left(\sum_{cyc}x_i\right)^{\frac{3}{2}}\geq\sum_{i=1}^nx^{\frac{3}{2}},$$ which follows from Karamata for a convex function $$f(x)=x^{\frac{3}{2}}.$$
Indeed, let $$x_1\geq x_2\geq...\geq x_n$$.
Thus, $$(x_1+x_2+...+x_n,0,...,0)\succ(x_1,x_2,...,x_n)$$ and we have: $$f(x_1+x_2+...+x_n)+f(0)+...+f(0)\geq f(x_1)+f(x_2)+...+f(x_n),$$ which is exactly that we need to prove.