Inequality, probably AM-GM inequality

Question

Prove that for $\{ x,y \in\Bbb R_+\ \}$
$\sqrt[3]{4x+4y} \ge \sqrt[3]{x} +\sqrt[3]{y} $

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Using the AM-GM inequality $ \sqrt[3]{x} +\sqrt[3]{y} \ge 2\sqrt[6]{xy} $ But I think I can't use it to prove the first inequality.

Answer 1

It's Holder: $$\sqrt[3]{4(x+y)}=\sqrt[3]{(1+1)(1+1)(x+y)}\geq\sqrt[3]{(\sqrt[3]x+\sqrt[3]y)^3}=\sqrt[3]x+\sqrt[3]y.$$

Also, it's the Power Mean inequality.

Indeed, let $x=a^3$ and $y=b^3$.

Hence, we need to prove that: $$\frac{a^3+b^3}{2}\geq\left(\frac{a+b}{2}\right)^3.$$

Now, we see that it's also the Jensen's inequality for $f(x)=x^3$.

The last inequality we can prove also by AM-GM.

Answer 2

Write $a=\sqrt[3]{x}, b= \sqrt[3]{y} $, so we have to prove $$4a^3+4b^3 \geq (a+b)^3$$

but this is the same as $$ 4(a^2-ab+b^2)\geq a^2+2ab+b^2$$

or $$ 3(a-b)^2\geq 0$$ which is true and no need for some fancy Holder inequality, or Power mean or Jensen inequality.