已知 $\alpha,\beta\in\left(0,\dfrac{\pi}2\right)$，求 $\dfrac{2\sin^4\alpha+3\cos^4\beta}{4\sin^2\alpha+5\cos^2\beta}+\dfrac{2\cos^4\alpha+3\sin^4\beta}{4\cos^2\alpha+5\sin^2\beta}$ 的最小值．
答案 $\dfrac 59$．
解析 设题中代数式为 $m$，则根据柯西不等式，有\[\begin{split} m&=\dfrac{2\sin^4\alpha}{4\sin^2\alpha+5\cos^2\beta}+\dfrac{2\cos^4\alpha}{4\cos^2\alpha+5\sin^2\beta}+\dfrac{3\cos^4\beta}{4\sin^2\alpha+5\cos^2\beta}+\dfrac{3\sin^4\beta}{4\cos^2\alpha+5\sin^2\beta}\\ &\geqslant \dfrac{\left(\sqrt 2\sin^2\alpha+\sqrt2\cos^2\alpha\right)^2+\left(\sqrt 3\cos^2\beta+\sqrt 3\sin^2\beta\right)^2}{(4\sin^2\alpha+5\cos^2\beta)+(4\cos^2\alpha+5\sin^2\beta)}\\ &=\dfrac 59,\end{split}\]等号当\[\begin{cases} \dfrac{\sqrt 2\sin^2\alpha}{4\sin^2\alpha+5\cos^2\beta}=\dfrac{\sqrt 2\cos^2\alpha}{4\cos^2\beta+5\sin^2\beta},\\ \dfrac{\sqrt 3\cos^2\beta}{4\sin^2\alpha+5\cos^2\beta}=\dfrac{\sqrt 3\sin^2\beta}{4\cos^2\beta+5\sin^2\beta},\end{cases}\iff \alpha+\beta=\dfrac{\pi}2\]时取得，因此所求最小值为 $\dfrac 59$．