Sempifying an algebraic expression

Question

I have the following expressions:$$f_1=a_1^2+a_2^2+\cdots+a_n^2$$ $$f_0=(a_1x_1+a_2x_2+\cdots+a_nx_n)^2$$ $$x_1^2+x_2^2+\cdots+x_n^2=1$$ I need to show that:$$f_1-f_0\ge0$$ for all $x_i$. Assume that all not the $a_i$'s are zero.

Answer 1

$$\sqrt{f_0}=\left|\sum_{i=1}^n a_ix_i\right|\underbrace{\le}_{\mathrm{Cauchy-Schwarz}} \sqrt{\sum_{i=1}^n a_i^2}\sqrt{\sum_{i=1}^n x_i^2}\underbrace{=}_{\sum_{i=1}^nx_i^2=1} \sqrt{\sum_{i=1}^n a_i^2}\implies f_0\le f_1.$$