# Proving Cauchy-Schwarz: $x_{1}y_{1}+x_{2}y_{2}\leq \sqrt{x_{1}^{2}+x_{2}^{2}} \sqrt{y_{1}^{2}+y_{2}^{2}}$

This is for a first year calculus course. Everything I can find online about Cauchy-Schwarz inequalities involves real analysis and vectors etc. I've only just begun calculus.

$x_1$, $x_2$, $y_1$, and $y_2$ are all real numbers.

Prove the Cauchy-Schwarz inequality: $$x_{1}y_{1}+x_{2}y_{2}\leq \sqrt{x_{1}^{2}+x_{2}^{2}} \sqrt{y_{1}^{2}+y_{2}^{2}}.$$

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Hint: square both sides, regroup, and compare with the expression $(x_1 y_2 - x_2 y_1)^2$. – Dan Brumleve Sep 14 '11 at 13:09
Look this page fatosmatematicos.blogspot.com/2009/08/… – Paulo Sérgio Sep 14 '11 at 13:10
you might take the absolute value of the right hand-side. Before it is too late :) – user13838 Sep 14 '11 at 15:14

Consider $P(t)=(x_1t-y_1)^2+(x_2t-y_2)^2$, where $x_1,x_2,y_1,y_2,t$ are all real. Clearly $P(t)\ge 0$ for all $t\in\mathbb{R}$. Since $P(t)$ can also be written as $$P(t)=(x_1^2+x_2^2)t^2-2(x_1y_1+x_2y_2)t+(y_1^2+y_2^2),$$ its discriminant must be smaller than or equal to $0$: $$D/4=(x_1y_1+x_2y_2)^2-(x_1^2+x_2^2)(y_1^2+y_2^2)\le 0.$$ Hence the Cauchy-Schwarz inequality follows.
So here's an really easy way to prove the inequality, involving only simply algebra: $$(x_{1}y_{2}-x_{2}y_{1})^2 \geq 0$$ $$x_{1}^2y_{2}^2- 2x_{1}y_{2}x_{2}y_{1}+ x_{2}^2y_{1}^2 \geq 0$$ $$x_{1}^2y_{2}^2+ x_{2}^2y_{1}^2 \geq 2x_{1}y_{2}x_{2}y_{1}$$ (we can actually jump straight to here by arithmetic mean greater than geometric mean if you are familiar with it) $$x_{1}^2y_{1}^2+ x_{2}^2y_{2}^2 + x_{1}^2y_{2}^2+ x_{2}^2y_{1}^2 \geq x_{1}^2y_{1}^2 + x_{2}^2y_{2}^2+ 2x_{1}y_{2}x_{2}y_{1}$$ $$(x_{1}^{2}+x_{2}^{2})(y_{1}^{2}+y_{2}^{2})\geq (x_{1}y_{1}+x_{2}y_{2})^2$$ since they are all positive, take square root and switch side to get: $$x_{1}y_{1}+x_{2}y_{2}\leq \sqrt{x_{1}^{2}+x_{2}^{2}} \sqrt{y_{1}^{2}+y_{2}^{2}}.$$