I've got stuck at this problem :
Prove that for any $a > 0$ and any $b > 0$ the following inequality is true: $$ {3} {\left(\frac{a^3}{b^3} + \frac{b^3}{a^3}\right)} \geq \frac{a}{b} +\frac{b}{a} + 4$$
The first thing that I've thought was the AM-GM inequality (the extended version - heard that is also known as The power mean inequality): $$ HM \leq GM \leq AM \leq SM $$ where $HM$, $GM$, $AM$, and $SM$ refer to the harmonic, geometric, arithmetic, and square mean, respectively. CBS(Cauchy - Buniakowsky - Schwartz) also come to my mind, but I think it isn't helpful in this case.
I would be greatful for some hints.
Thanks!
$\endgroup$ 12 Answers
$\begingroup$Let $ x = \frac{a}b+\frac{b}a \ge 2$ (by AM-GM). Then $$x^3=\frac{a^3}{b^3}+\frac{b^3}{a^3} + 3x$$ So we want to show, for $x \ge 2$, $$3(x^3-3x) \ge x+4 \iff (x-2)(3x^2+6x+2) \ge 0$$ which is obvious.
$\endgroup$ $\begingroup$this inequation is equivalent to $$\frac{(a-b)^2 \left(3 a^4+6 a^3 b+8 a^2 b^2+6 a b^3+3 b^4\right)}{a^3 b^3}\geq 0$$ which is true.
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