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: $$ x_{1}y_{1}+x_{2}y_{2}\leq \sqrt{x_{1}^{2}+x_{2}^{2}} \sqrt{y_{1}^{2}+y_{2}^{2}}. $$
 A: 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.
A: 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:
$$ x_{1}y_{1}+x_{2}y_{2}\leq \sqrt{x_{1}^{2}+x_{2}^{2}} \sqrt{y_{1}^{2}+y_{2}^{2}}. $$
