Prove $\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$

Question

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?

Answer 1

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 $(a-b)^2 + (a-c)^2 + (a-d)^2 + (b-c)^2 + (b-d)^2 + (c-d)^2 \ge 0$

Answer 2

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