For positive real numbers $a,b$ prove that $\sqrt[3]{2(a+b)(\frac{1}{a}+\frac{1}{b})}\ge\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}$.

Question

For positive real numbers $a,b$ prove that $$\sqrt[3]{2(a+b)(\frac{1}{a}+\frac{1}{b})}\ge\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}$$

Here's what I've gotten to so far,
$2(a+b)(\frac{1}{a}+\frac{1}{b})$
$=2(1+\frac{a}{b}+\frac{b}{a}+1)$
$=2(2+\frac{a^2+b^2}{ab})$
$=2(\frac{a^2+2ab+b^2}{ab})$
$=2(\frac{(a+b)^2}{ab})$
I'm not sure where to go from here, I'm pretty sure I have to use the $A.M-G.M$ inequality in some way, but I'm not sure how it'd be used as both sides of the inequality look like $G.M$ inequalities. Any help is appreciated.

Answer 1

Hint:

Let $x=\sqrt[3]{a/b}$, then you have to prove $$4+2x^3+{2\over x^3}\geq (x+{1\over x})^3$$

Notice that if $t>0$ then $$t^3-3t+2 = (t+2)(t-1)^2\geq 0 $$

Answer 2

Starting from Aqua answer, with $x = \sqrt[3]{b\over a}\;$, but then let $y=\left(x+{1\over x}\right)$

$y^3 = \left(x+{1\over x}\right)^3 = \left(x^3+{1\over x^3}\right) + 3y\;$

We have to proof:

$\begin{align} 4 + 2·(y^3 - 3y) &≥ y^3\\ y^3 - 6y + 4 &≥ 0 \end{align}$

AM-GM inequality: $\;y = \left(x+\frac{1}{x}\right) ≥ 2×\sqrt{x×\frac{1}{x}} = 2\;$
Let $z=y-2$

$(z+2)^3 - 6(z+2) + 4 = (z^3+6z^2+12z+8) - 6z - 8 = z^3+6z^2+6z ≥ 0$

With positive coefficients, and $z≥0$, above is always true. QED