已知 $x,y\in (0,1)$，求证：\[\left|x\sin\dfrac{1}{x^2}-y\sin\dfrac{1}{y^2}\right|\leqslant 3\sqrt[3]{|x-y|}.\]
解析 不妨设 $x\geqslant y$，根据题意，有\[\begin{split} \left|x\sin\dfrac{1}{x^2}-y\sin\dfrac{1}{y^2}\right|&= \left|(x-y) \sin \frac{1}{x^2}+y\left(\sin \frac{1}{x^2}-\sin \frac{1}{y^2}\right)\right| \\ & \leqslant |x-y| \cdot\left|\sin \frac{1}{x^2}\right|+2|y| \cdot\left|\sin \frac{x^2-y^2}{2 x^2 y^2}\right| \\ & \leqslant (x-y)+2y \cdot \min \left\{1, \frac{x^2-y^2}{2 x^2 y^2}\right\} \\ & =(x-y)+\min \left\{2 y, \frac{x^2-y^2}{x^2 y}\right\} \\ & \leqslant \sqrt[3]{x-y}+\sqrt[3]{2 y \cdot 2 y \cdot \frac{x^2-y^2}{x^2 y}} \\ & =\sqrt[3]{x-y}+\sqrt[3]{4(x-y) \cdot \frac{(x+y) y}{x^2}} \\ & \leqslant \sqrt[3]{x-y}+\sqrt[3]{4(x-y) \cdot \frac{2 x^2}{x^2}} \\ & =3 \sqrt[3]{x-y} . \end{split}\]