<s>[INST] Another inequality to prove involving arbitrary reals Let $a,b$ be non-zero reals such that $ab\ge \frac{1}{a}+\frac{1}{b}+3$ then prove the following inequality :
$$ \sqrt[3]{ab}\ge \frac{1}{\sqrt[3]{a}}+\frac{1}{\sqrt[3]{b}}$$
This one stumped me completely as usual. A solution would be welcome.
 [/INST] I found another proof that establishes the equivalence between the two inequalities:
[As in my other solution, I will substitute all the variables with their cubes.]
$a^3 b^3 - \frac{1}{a^3}-\frac{1}{b^3} - 3 = ( a b - \frac{1}{a} - \frac{1}{b} ) ( a^2 b^2 + a + b + \frac{1}{a^2} + \frac{1}{b^2} - \frac{1}{ab} )$
$ = ( a b - \frac{1}{a} - \frac{1}{b} ) \left( \frac{1}{2}(ab+\frac{1}{a})^2 + \frac{1}{2}(ab+\frac{1}{b})^2+\frac{1}{2}(\frac{1}{a}-\frac{1}{b})^2 \right)$
Also $\frac{1}{2}(ab+\frac{1}{a})^2 + \frac{1}{2}(ab+\frac{1}{b})^2+\frac{1}{2}(\frac{1}{a}-\frac{1}{b})^2 \ge 0$ with equality case only when $a = b = -1$
And $a b - \frac{1}{a} - \frac{1}{b} = 3 > 0$ when $a = b = -1$
Therefore $a^3 b^3 - \frac{1}{a^3}-\frac{1}{b^3} - 3 \ge 0$ iff $a b - \frac{1}{a} - \frac{1}{b} \ge 0$
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