How to prove Cauchy-Schwarz Inequality in $R^3$?

I am having trouble proving this inequality in $R^3$. It makes sense in $R^2$ for the most part. Can anyone at least give me a starting point to try. I am lost on this thanks in advance.

Answer

You know that, for any $x,y$, we have that

$$(x-y)^2\geq 0$$

Thus

$$y^2+x^2\geq 2xy$$

Cauchy-Schwarz states that

$$x_1y_1+x_2y_2+x_3y_3\leq \sqrt{x_1^2+x_2^2+x_3^3}\sqrt{y_1^2+y_2^2+y_3^3}$$

Now, for each $i=1,2,3$, set

$$x=\frac{x_i}{\sqrt{x_1^2+x_2^2+x_3^2}}$$

$$y=\frac{y_i}{\sqrt{y_1^2+y_2^2+y_3^2}}$$

We get

$$\frac{y_1^2}{{y_1^2+y_2^2+y_3^2}}+\frac{x_1^2}{{x_1^2+x_2^2+x_3^2}}\geq 2\frac{x_1}{\sqrt{x_1^2+x_2^2+x_3^2}}\frac{y_1}{\sqrt{y_1^2+y_2^2+y_3^2}}$$

$$\frac{y_2^2}{{y_1^2+y_2^2+y_3^2}}+\frac{x_2^2}{{x_1^2+x_2^2+x_3^2}}\geq 2\frac{x_2}{\sqrt{x_1^2+x_2^2+x_3^2}}\frac{y_2}{\sqrt{y_1^2+y_2^2+y_3^2}}$$

$$\frac{y_3^2}{{y_1^2+y_2^2+y_3^2}}+\frac{x_3^2}{{x_1^2+x_2^2+x_3^2}}\geq 2\frac{x_3}{\sqrt{x_1^2+x_2^2+x_3^2}}\frac{y_3}{\sqrt{y_1^2+y_2^2+y_3^2}}$$

Summing all these up, we get

$$\frac{y_1^2+y_2^2+y_3^2}{{y_1^2+y_2^2+y_3^2}}+\frac{x_1^2+x_2^2+x_3^2}{{x_1^2+x_2^2+x_3^2}}\geq 2\frac{x_1y_1+x_2y_2+x_3y_3}{\sqrt{y_1^2+y_2^2+y_3^2}\sqrt{x_1^2+x_2^2+x_3^2}}$$

$$\sqrt{y_1^2+y_2^2+y_3^2}\sqrt{x_1^2+x_2^2+x_3^2}\geq {x_1y_1+x_2y_2+x_3y_3}$$

This works for $\mathbb R^n$. We sum up through $i=1,\dots,n$ and set

$$y=\frac{y_i}{\sqrt{\sum y_i^2}}$$

$$x=\frac{x_i}{\sqrt{\sum x_i^2}}$$

Note this stems from the most fundamental inequality $x^2\geq 0$.