Giải thích các bước giải: ta có
$\begin{array}{l}
\\
{a^4} + {b^4} = {({a^2} + {b^2})^2} - 2{a^2}{b^2}\\
= {\left[ {{{(a + b)}^2} - 2ab} \right]^2} - 2{a^2}{b^2}\\
= {(1 - 2ab)^2} - 2{a^2}{b^2}\\
= 1 - 4ab + 4{a^2}{b^2} - 2{a^2}{b^2}\\
= 2{a^2}{b^2} - 4ab + 1\\
= 2({a^2}{b^2} - 2ab + 1) - 1\\
= 2{(ab - 1)^2} - 1\\
a + b \ge 2\sqrt {ab} \\
< = > 1 \ge 2\sqrt {ab} \\
< = > \sqrt {ab} \le \frac{1}{2}\\
< = > 0 \le ab \le \frac{1}{4}\\
< = > - 1 \le ab - 1 \le - \frac{3}{4}\\
< = > \frac{9}{{16}} \le {(ab - 1)^2} \le 1\\
< = > 2.\frac{9}{{16}} \le 2{(ab - 1)^2} \le 2.1\\
< = > \frac{9}{8} - 1 \le 2{(ab - 1)^2} - 1 \le 2 - 1\\
< = > \frac{1}{8} \le 2{(ab - 1)^2} - 1 \le 1\\
= > {a^4} + {b^4} \ge \frac{1}{8}(đpcm)
\end{array}$
