a) Ta có:
$\dfrac{a^3}{a^2 + b^2} = \dfrac{a^3 + ab^2 - ab^2}{a^2 + b^2} = \dfrac{a(a^2 + b^2) - ab^2}{a^2 + b^2} = a - \dfrac{ab^2}{a^2 + b^2}$
Ta lại có:
$a^2 + b^2 \geq 2\sqrt{a^2b^2}$
$\Rightarrow \dfrac{ab^2}{a^2 + b^2} \leq \dfrac{ab^2}{2\sqrt{a^2b^2}}$
$\Rightarrow a - \dfrac{ab^2}{a^2 + b^2} \geq a - \dfrac{ab^2}{2\sqrt{a^2b^2}} = a - \dfrac{b}{2}$
$\Rightarrow \dfrac{a^3}{a^2 + b^2} \geq a - \dfrac{b}{2}$
Chứng minh tương tự, ta được:
$\dfrac{b^3}{b^2 + c^2} \geq b - \dfrac{c}{2}$
$\dfrac{c^3}{c^2 + d^2} \geq c - \dfrac{d}{2}$
$\dfrac{d^3}{d^2 + a^2} \geq d - \dfrac{a}{2}$
Cộng vế theo vế, ta được:
$\dfrac{a^3}{a^2 + b^2} + \dfrac{b^3}{b^2 + c^2} + \dfrac{c^3}{c^2 + d^2} + \dfrac{d^3}{d^2 + a^2} \geq a - \dfrac{b}{2} + b - \dfrac{c}{2}+c - \dfrac{d}{2}+d - \dfrac{a}{2} = \dfrac{a +b + c +d}{2}$
Dấu = xảy ra khi $a =b = c = d$
2) Ta có:
$\dfrac{a^4}{a^3 + 2b^3} = \dfrac{a^4 + 2ab^3 - 2ab^3}{a^3 + 2b^3} = \dfrac{a(a^3 + 2b^3)}{a^3 + 2b^3} - \dfrac{2ab^3}{a^3 + 2b^3} = a - \dfrac{2ab^3}{a^3 + 2b^3}$
Ta lại có:
$a^3 +2b^3 = a^3 + b^3 + b^3 \geq 3\sqrt[3]{a^3b^3b^3} = 3ab^2$
$\Rightarrow \dfrac{2ab^3}{a^3 + 2b^3} \leq \dfrac{2ab^3}{3ab^2} = \dfrac{2b}{3}$
$\Rightarrow a - \dfrac{2ab^3}{a^3 + 2b^3} \geq a - \dfrac{2b}{3}$
$\Rightarrow \dfrac{a^4}{a^3 + 2b^3} \geq a - \dfrac{2b}{3}$
Chứng minh tương tự, ta được:
$\dfrac{b^4}{b^3 + 2c^3} \geq b - \dfrac{2c}{3}$
$\dfrac{c^4}{c^3 + 2d^3} \geq c - \dfrac{2d}{3}$
$\dfrac{d^4}{d^3 + 2a^3} \geq d - \dfrac{2a}{3}$
Cộng vế theo vế, ta đươc:
$ \dfrac{a^4}{a^3 + 2b^3} + \dfrac{b^4}{b^3 + 2c^3} + \dfrac{c^4}{c^3 + 2d^3} + \dfrac{d^4}{d^3 + 2a^3} \geq a - \dfrac{2b}{3} + b - \dfrac{2c}{3} + c - \dfrac{2d}{3} + d - \dfrac{2a}{3} = \dfrac{a + b + c + d}{3}$
Dấu = xảy ra khi $a = b = c =d$
