<s>[INST] 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?
 [/INST] Use this inequality:
$\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab$ equivalent $\frac{1}{4}(a+b+c+d) \geq \sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$ Now just prove$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq \frac{1}{4}(a+b+c+d)$ equivalent $4(a^2+b^2+c^2+d^2)\geq (a+b+c+d)^2$ equivalent $3(a^2+b^2+c^2+d^2)\geq 2(ab+ac+ad+ bc + bd + cd)$ equivalent $(a-b)^2 + (a-c)^2 + (a-d)^2 + (b-c)^2 + (b-d)^2  + (c-d)^2 \ge 0$
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