Giải thích các bước giải:
Ta có:
$(\dfrac{x^4}{a}+\dfrac{y^4}{b})(a+b)\ge (\sqrt{\dfrac{x^4}{a}\cdot a}+\sqrt{\dfrac{y^4}{b}\cdot b})^2$ (BĐT bunhiaxcopki)
$\to (\dfrac{x^4}{a}+\dfrac{y^4}{b})(a+b)\ge (x^2+y^2)^2$
$\to (\dfrac{x^4}{a}+\dfrac{y^4}{b})(a+b)\ge 1$ vì $x^2+y^2=1$
$\to \dfrac{x^4}{a}+\dfrac{y^4}{b}\ge \dfrac{1}{a+b}$
Dấu = xảy ra khi:
$\dfrac{\dfrac{x^2}{\sqrt{a}}}{\sqrt{a}}=\dfrac{\dfrac{y^2}{\sqrt{b}}}{\sqrt{b}}$
$\to \dfrac{x^2}{a}=\dfrac{y^2}{b}$
$\to \dfrac{x}{\sqrt{a}}=\dfrac{y}{\sqrt{b}}$
$\to \dfrac{x}{\sqrt{a}}+\dfrac{\sqrt{b}}{y}=\dfrac{y}{\sqrt{b}}+\dfrac{\sqrt{b}}{y}$
$\to \dfrac{x}{\sqrt{a}}+\dfrac{\sqrt{b}}{y}\ge 2\sqrt{\dfrac{y}{\sqrt{b}}\cdot\dfrac{\sqrt{b}}{y}}$
$\to \dfrac{x}{\sqrt{a}}+\dfrac{\sqrt{b}}{y}\ge 2$
