Lời giải:
Áp dụng BĐT AM-GM:
\(1\geq a+b\geq 2\sqrt{ab}\Rightarrow ab\leq \frac{1}{4}\)
\(\frac{a}{2}+\frac{a}{2}+\frac{1}{16a^2}\geq 3\sqrt[3]{\frac{a}{2}.\frac{a}{2}.\frac{1}{16a^2}}=\frac{3}{4}(1)\)
\(\frac{b}{2}+\frac{b}{2}+\frac{1}{16b^2}\geq 3\sqrt[3]{\frac{b}{2}.\frac{b}{2}.\frac{1}{16b^2}}=\frac{3}{4}(2)\)
\(\frac{15}{16}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\geq \frac{15}{16}.2\sqrt{\frac{1}{a^2}.\frac{1}{b^2}}=\frac{15}{8ab}\geq \frac{15}{8.\frac{1}{4}}=\frac{15}{2}(3)\)
Lấy \((1)+(2)+(3)\Rightarrow a+b+\frac{1}{a^2}+\frac{1}{b^2}\geq \frac{3}{4}+\frac{3}{4}+\frac{15}{2}=9\) (đpcm)
Dấu "=" xảy ra khi $a=b=\frac{1}{2}$
