南京大学 2023年强基第4题

1 解：由柯西不等式可知 $\displaystyle \frac{\sin ^{4} \alpha}{\sin ^{2} \beta}+\frac{\cos ^{4} \alpha}{\cos ^{2} \beta}=1 \geqslant \frac{\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}}{\sin ^{2} \beta+\cos ^{2} \beta}=\frac{1}{1}=1$ 当 $\displaystyle \frac{\sin ^{2} \alpha}{|\sin \beta|}=\frac{\cos ^{2} \alpha}{|\cos \beta|}=\frac{\sqrt{2}}{2}$ 时，等号成立，此时 $\displaystyle \sin ^{2} \alpha=\frac{\sqrt{2}}{2}|\sin \beta|, \cos ^{2} \alpha=\frac{\sqrt{2}}{2}|\cos \beta|$ 则 $\displaystyle \frac{\sqrt{2}}{2}(|\sin \beta|+|\cos \beta|)=1 \Rightarrow|\sin \beta|+|\cos \beta|=\sqrt{2} \Rightarrow|\sin \beta \cos \beta|=\frac{1}{2}$ 于是 $\displaystyle \frac{\sin ^{4} \beta}{\sin ^{2} \alpha}+\frac{\cos ^{4} \beta}{\cos ^{2} \alpha}=\sqrt{2}\left(|\sin \beta|^{3}+|\cos \beta|^{3}\right)$ $=\sqrt{2}(|\sin \beta|+|\cos \beta|)\left(\sin ^{2} \beta+\cos ^{2} \beta-|\sin \beta \cos \beta|\right)$ $\displaystyle =\sqrt{2} \times \sqrt{2} \times\left(1-\frac{1}{2}\right)=1$.