Let $x_i,y_i\in\mathbb{R},\quad i=1,\dots,n$. Show that $$(\sum_{i=1}^n x_iy_i)^2\leq \sum_{i=1}^n x_i^2\sum_{i=1}^n y_i^2$$ where the equality holds if and only if $x_i=\lambda y_i$ where $\lambda_i\in\mathbb{R}$
If I think I already saw it in some place but I don't remember how to proof
For $n=1$ $$(x_1y_1)^2=x_1^2y_1^2\leq x_1^2\times y_1^2$$
Assuming that is true for $n=k$ for any $k\in\mathbb{Z}$. Then $$(\sum_{i=1}^k x_iy_i)^2\leq \sum_{i=1}^k x_i^2\sum_{i=1}^k y_i^2$$ Then we need to show that for $n=k+1$ it is true too $$(\sum_{i=1}^{k+1} x_iy_i)^2\leq \sum_{i=1}^{k+1} x_i^2\sum_{i=1}^{k+1} y_i^2$$
I'm stuck now, I thought it was something related to moving the series but I got nowhere.