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.

This question is missing context or other details: Please improve the question by providing additional context, which ideally includes the source and motivation of the problem, your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. — Carl Mummert, Jul 30 '18 at 19:26
I think this question does not miss context and in this question just there are details. — Michael Rozenberg, Jul 30 '18 at 19:32
Possible duplicate of How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}}$ — Isaac Browne, Jul 30 '18 at 21:06

score 0 · Accepted Answer · answered Jul 30 '18 at 19:31

0

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

answered Jul 30 '18 at 19:31

Michael Rozenberg

194,933

Why down voted? – Michael Rozenberg Jul 30 '18 at 21:05

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

1 Answers1