Let $x_1, x_2, \dots x_k \ge 0$ be non-negative real numbers.
Does it follow that $$k \left( \sum_{i=1}^k x_i^3 \right)^2 \ge \left( \sum_{i=1}^k x_i^2 \right)^3 ? $$
This seems like something that might easily follow from standard inequalities like Jensen's inequality?
(Now I am embarrassed that I hadn't really carefully tried Jensen before asking. As a penance, I will post a solution using Jensen's inequality.)
On
Using Holder, $$\sum_{i=1}^kx_i^2\leq k^{\frac{1}{3}}\left(\sum_{i=1}^kx_i^3\right)^{\frac{2}{3}}.$$
Jensen also works with the convex function $x\mapsto x^\frac{3}{2}$, what gives $$\left(\frac{1}{k}\sum_{i=1}^k x_i^2\right)^{\frac{3}{2}}\leq \frac{1}{k}\sum_{i=1}^kx_i^3.$$
On
Also, we can use Jensen here.
Indeed, let $x_i^2=a_i$. Thus, by Jensen for the convex function $f(x)=x^{\frac{3}{2}}$ we obtain: $$\frac{\sum\limits_{i=1}^ka_i^{\frac{3}{2}}}{k}\geq\left(\frac{\sum\limits_{i=1}^ka_i}{k}\right)^{\frac{3}{2}},$$ which is our inequality exactly!
On
In case it is helpful for anyone else, I will post another solution using Jensen's inequality.
Jensen's inequality says that for a convex function $\phi$, numbers in its domain $y_1, y_2, \dots, y_k$, and positive weights $a_1, a_2, \dots, a_k$, we have $$\phi \left( \frac{\sum_{i=1}^k a_i y_i }{\sum_{i=1}^k a_i} \right) \le \frac{\sum_{i=1}^k a_i \phi(y_i) }{\sum_{i=1}^k a_i}.$$
Set $a_i = 1$ and $y_i = x_i^2$ for $i=1, 2, \dots k $, and let $\phi(x) = x^{3/2}$. Then we have $$\left( \frac{\sum_{i=1}^k x_i^2 }{k} \right)^{3/2} \le \frac{\sum_{i=1}^k x_i^3 }{k}.$$ Squaring both sides gives the desired result.
By Holder $$k\left(\sum_{i=1}^kx_i^3\right)^2=\sum_{i=1}^k1\left(\sum_{i=1}^kx_i^3\right)^2\geq\left(\sum_{i=1}^k\sqrt[3]{1\cdot\left(x_i^3\right)^2}\right)^{1+2}=\left(\sum_{i=1}^kx_i^2\right)^3.$$