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
$$(sum_{i}^{n}x_{i}y_{i})^{2}leq sum_{i}^{n}x_{i}^{2}sum_{i}^{n}y_{i}^{2}$$