Ta có: $a^4+b^4=(a^4+\frac{1}{16}+\frac{1}{16}+\frac{1}{16})+(b^4+\frac{1}{16}+\frac{1}{16}+\frac{1}{16})-\frac{3}{8}$
$\geq 4.\sqrt[4]{a^4.\frac{1}{16}.\frac{1}{16}.\frac{1}{16}}+4.\sqrt[4]{b^4.\frac{1}{16}.\frac{1}{16}.\frac{1}{16}}-\frac{3}{8}$
$=\frac{a+b}{2}-\frac{3}{8}>\frac{1}{2}-\frac{3}{8}=\frac{1}{8}.\Rightarrow $ đpcm.
Ấn dấu tick nếu đáp án đúng.