Đáp án:
$Min(a^4+b^4)=\dfrac{1}{8}$
Giải thích các bước giải:
Ta có:
$\begin{cases}a^4+\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}\ge 4\sqrt[4]{a^4.\dfrac{1}{16}.\dfrac{1}{16}.\dfrac{1}{16}}=\dfrac{a}{2}\\ b^4+\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}\ge 4\sqrt[4]{b^4.\dfrac{1}{16}.\dfrac{1}{16}.\dfrac{1}{16}}=\dfrac{b}{2}\end{cases}$
$\rightarrow (a^4+\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16})+(b^4+\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16})\ge \dfrac{a}{2}+\dfrac{b}{2}$
$\rightarrow a^4+b^4+\dfrac{3}{8}\ge \dfrac{1}{2}$
$\rightarrow a^4+b^4\ge \dfrac{1}{8}$
Dấu = xảy ra khi $a=b=\dfrac{1}{2}$
