Prove Cauchy-Schwarz inequality for two n-dimensional vectors

Question

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"). — Angina Seng, 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

Olivier Oloa · Accepted Answer · 2017-11-25T13:00:03.570

2

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 $\Delta'=b^2-ac$?

edited Nov 25 '17 at 13:00

answered Nov 25 '17 at 12:35

Olivier Oloa

120,989

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

Prove Cauchy-Schwarz inequality for two n-dimensional vectors

1 Answers1