Prove $\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$ For real numbers $a, b, c, d >0$ I have to prove that  $$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$$ always holds. I tried using the mean inequalities for multiple numbers but I always get in trouble with the different orders of the two roots. And in case I make an estimate that can handle the orders of the roots, it is too strong. Can anyone help?

Since 
$$\frac{a^2+b^2+c^2+d^2}{4}\geq\frac{ab+ac+bc+ad+bd+cd}{6}$$
it's just
$$(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2\geq0,$$ it's enough to prove that
$$\sqrt{\frac{ab+ac+bc+ad+bd+cd}{6}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}},$$
which I proved here:
How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}}$
