Another inequality to prove involving arbitrary reals

Question

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.

Start by cubing both sides of the inequality you are trying to prove. — Lee Mosher, Jun 01 '14 at 15:20
It reverts back to the same inequality. Or maybe I did a mistake but I am getting the same result. Can you please clarify? — shadow10, Jun 01 '14 at 15:24
Note that the equality condition is $ a = b = 2$. (Not sure how to use it yet). — Calvin Lin, Jun 01 '14 at 15:53
Equality condition is in fact $b = \dfrac{3a+1 \pm (a+1)\sqrt{4a+1}}{2a^2}$, for e.g. $a=6, b =\frac34$ also gives inequality. — Macavity, Jun 02 '14 at 01:27

score 2 · Answer 1 · answered Jun 02 '14 at 03:18

[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.]

score 2 · Accepted Answer · answered Jun 02 '14 at 08:12

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 $a b - \frac{1}{a} - \frac{1}{b} \ge 0$

Another inequality to prove involving arbitrary reals

2 Answers2