EM THAM KHẢO :

Với $x > 0$, $f(x) = x(\pi - x^2)$ ta xét $\pi - x^2 > 0$

Áp dụng bất đẳng thức Cauchy:

$2x^2(\pi - x^2)(\pi - x^2) \le \left(\dfrac{2x^2 + \pi - x^2 + \pi - x^2}{3}\right)^3$

$\Leftrightarrow 2x^2(\pi - x^2)^2 \le \left(\dfrac{2\pi}{3}\right)^3$

$\Leftrightarrow 2x^2(\pi - x^2)^2 \le \dfrac{8\pi^3}{27}$

$\Leftrightarrow x^2(\pi - x^2)^2 \le \dfrac{4\pi^3}{27}$

$\Leftrightarrow x(\pi - x^2) \le \sqrt{\dfrac{4\pi^3}{27}}$

$\Leftrightarrow f(x) \le \dfrac{2\pi\sqrt{3\pi}}{9}$

Dấu "=" xảy ra $\Leftrightarrow 2x^2 = \pi - x^2 \Leftrightarrow 3x^2 = \pi \Leftrightarrow x = \sqrt{\dfrac{\pi}{3}}$

Từ điểm rơi $x = \sqrt{\dfrac{\pi}{3}}$, ta khai triển VP để CM:

$-\left(x - \sqrt{\dfrac{\pi}{3}}\right)^2 \left(x + 2\sqrt{\dfrac{\pi}{3}}\right) + \dfrac{2\pi\sqrt{3\pi}}{9} = -\left(x^2 - 2\sqrt{\dfrac{\pi}{3}}x + \dfrac{\pi}{3}\right)\left(x + 2\sqrt{\dfrac{\pi}{3}}\right) + \dfrac{2\pi\sqrt{3\pi}}{9}$

$= -\left(x^3 + 2\sqrt{\dfrac{\pi}{3}}x^2 - 2\sqrt{\dfrac{\pi}{3}}x^2 - \dfrac{4\pi}{3}x + \dfrac{\pi}{3}x + \dfrac{2\pi}{3}\sqrt{\dfrac{\pi}{3}}\right) + \dfrac{2\pi\sqrt{3\pi}}{9}$

$= -\left(x^3 - \pi x + \dfrac{2\pi\sqrt{3\pi}}{9}\right) + \dfrac{2\pi\sqrt{3\pi}}{9}$

$= -x^3 + \pi x - \dfrac{2\pi\sqrt{3\pi}}{9} + \dfrac{2\pi\sqrt{3\pi}}{9}$

$= -x^3 + \pi x$

$= x(\pi - x^2)$

$\Rightarrow f(x) = -\left(x - \sqrt{\dfrac{\pi}{3}}\right)^2 \left(x + 2\sqrt{\dfrac{\pi}{3}}\right) + \dfrac{2\pi\sqrt{3\pi}}{9}$

Với $\forall x > 0 \Rightarrow x + 2\sqrt{\dfrac{\pi}{3}} > 0$

$\Rightarrow -\left(x - \sqrt{\dfrac{\pi}{3}}\right)^2 \left(x + 2\sqrt{\dfrac{\pi}{3}}\right) \le 0$

$\Rightarrow f(x) \le \dfrac{2\pi\sqrt{3\pi}}{9}$