# Prove Cauchy-Schwarz inequality for two n-dimensional vectors

I have to prove that for any real numbers $x_1,x_2,..., x_n$ and $y_1,y_2,...,y_n$ we have: $$(|x_1y_1|+|x_2y_2|+...+|x_ny_n|)^2\le (x_1^2+x_2^2+...+x_n^2)(y_1^2+y_2^2+...+y_n^2)$$ I'm supposed to start from this identity: $$(|x_1|-t|y_1|)^2+...+(|x_n|-t|y_n|)^2 \ge 0\;\forall t$$ Could you give me some advice?

I also need to prove Minkowski's inequality starting from Cauchy-Schwarz: $$(\sum_{i=1}^n|x_i+y_i|^2)^{1/2}\le (\sum_{i=1}^n|x_i|^2)^{1/2} +(\sum_{i=1}^n |y_i|^2)^{1/2}$$

• Identify the coefficients of $t^2$, $t^1$ and $t^0$ on the LHS of your inequality (not "identity"). – Lord Shark the Unknown Nov 25 '17 at 12:34
• It may be helpful to notice that inequality can be expressed as $|x\cdot y| \le \|x\|\|y\|$ where $x=(|x_1|,\dots,|x_n|)$ and similarly for $y$. – user160738 Nov 25 '17 at 13:05
• Possible duplicate of Proof of Cauchy-Schwarz Inequality – Martin R Nov 25 '17 at 13:08

## 1 Answer

Hint. By expanding your last expression, one obtains a non-negative quadratic polynomial: $$at^2+2bt+c\ge0, \qquad t\in \mathbb{R},$$ what can then be said about the sign of $\Delta'=b^2-ac$?

• No idea. I'm not sure about how to square a summation. – Seven Nov 25 '17 at 16:13
• Recall that $(a-b)^2=a^2-2ab+b^2$, then apply it to $(|x_1|-t|y_1|)^2$. Do the same for each square $(|x_k|-t|y_k|)^2$, then collect the results... – Olivier Oloa Nov 25 '17 at 16:50