柯西-许瓦兹尔不等式

设 $\lambda_{i}=\frac{y_{i}^{2}}{\sum_{i}^{n} y_{i}^{2}}$, 则 $\sum_{i}^{n}\lambda_{i}=1$.
考虑函数 $f(x)=x^2$ , $f''(x)=2>0$,利用凸不等式
$$f(\sum_{i}^{n}\lambda_{i}\frac{x_{i}}{y_{i}})\leq \sum_{i}^{n}\lambda_{i}f(\frac{x_{i}}{y_{i}})$$
整理下可得
$$(\frac{\sum_{i}^{n}x_{i}y_{i}}{\sum_{i}^{n}y_{i}^{2}})^{2}\leq \frac{\sum_{i}^{n}x_{i}^{2}}{\sum_{i}^{n}y_{i}^{2}}$$
即是
$$(\sum_{i}^{n}x_{i}y_{i})^{2}\leq \sum_{i}^{n}x_{i}^{2}\sum_{i}^{n}y_{i}^{2}$$