Prove that $\sqrt{\frac{a^2+b^2+c^2+d^2}{4}} \ge \sqrt[3]{\frac{abc+bcd+cda+dab}{4}}$.

Question

Let $a$, $b$, $c$ and $d$ be some positive integers. Prove that:

$$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}} \ge \sqrt[3]{\frac{abc+bcd+cda+dab}{4}}$$

I was trying to use $A.M. \ge G.M.$, but it's not enough:

$$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}} \ge \sqrt[4]{abcd}$$

Can you help me, please? Thanks!

Edit: I have just now seen that this question is a duplicate. I didn't saw it in the list of suggested questions when I was writing it. Sorry.

Answer 1

By AM-GM and Maclaurin we obtain: $$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt{\frac{ab+ac+ad+bc+bd+cd}{6}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}.$$ A proof of the second inequality see here:

How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}}$