1615

若 ${\displaystyle a>0,b>0}$, ${\displaystyle \frac{3}{a}+\frac{4}{b}=1}$, 求 ${\displaystyle \frac{1}{a^2}+\frac{1}{b^2}}$ 的最小值.

代换 ${\displaystyle m=\frac{3}{a},n=\frac{4}{b}}$, 则根据柯西不等式 $$(\frac{m^2}{9}+\frac{n^2}{16})(9+16)\ge (m+n{)}^2=1,$$ 从而 ${\displaystyle \mathrm{原}\mathrm{式}=\frac{m^2}{9}+\frac{n^2}{16}\ge \frac{1}{25}}$.