<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] [Clearly we can substitute all the variables with their cubes, which I will do so to avoid writing so many cube-roots.]
For any $a,b < 0$:
  $a b > 0 > \frac{1}{a} + \frac{1}{b}$
For any $a > 0$ and $b \ne 0$:
  If $a b < \frac{1}{a} + \frac{1}{b}$:
    Let $b' > b$ such that $a b' = \frac{1}{a} + \frac{1}{b'}$ and ( $b' < 0$ if $b < 0$ ), which is possible because:
      $f : x \mapsto a x - \frac{1}{a} - \frac{1}{x}$ is continuous except at $0$
      $f(b) < 0$ and $f(0^-) = \infty$ and $f(\infty) = \infty$
      Thus the claim follows by Intermediate value theorem
    Then $a^3 b^3 < a^3 b'^3 = ( \frac{1}{a} + \frac{1}{b'} )^3 = \frac{1}{a^3} + \frac{1}{b'^3} + 3 \frac{1}{ab'}( \frac{1}{a} + \frac{1}{b'} ) = \frac{1}{a^3} + \frac{1}{b'^3} + 3 < \frac{1}{a^3} + \frac{1}{b^3} + 3$
  Therefore $a b \ge \frac{1}{a} + \frac{1}{b}$ if $a^3 b^3 \ge \frac{1}{a^3} + \frac{1}{b^3} + 3$
Likewise for $b > 0$ and $a \ne 0$
Therefore ( $a b \ge \frac{1}{a} + \frac{1}{b}$ if $a^3 b^3 \ge \frac{1}{a^3} + \frac{1}{b^3} + 3$ ) for any $a,b \ne 0$
[I should add that this solution was found by plotting the two inequalities using Graph and seeing that they were equivalent. The solution is essentially to show that equality for the simpler one implies equality for the other, and then to move within the regions of inequality towards the equality case.]
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