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.
 A: 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 $x^2\geq 0$.
A: For any $x_i, x_j, y_i, y_j$ we have $$(x_i y_j - x_j y_i)^2 = x_i^2 y_j^2 - 2 x_i x_j y_i y_j + x_j^2 y_i^2 \geq 0$$ which gives $$x_i^2 y_j^2  + x_j^2 y_i^2 \geq 2 x_i x_j y_i y_j.$$
Then for $x,y \in \mathbb{R}^n$ (ie, not just $\mathbb{R}^3$):
$$2(\sum_{i=1}^n x_i y_i)^2 = 2\sum_{i=1}^n \sum_{j=1}^n x_i x_j y_i y_j  \leq \sum_{i=1}^n \sum_{j=1}^n (x_i^2 y_j^2  + x_j^2 y_i^2) = 2(\sum_{i=1}^n x_i^2)(\sum_{i=1}^n y_i^2)$$
Dividing by $2$ and taking square roots gives the desired result.
A: Let $f(t)$ be  the square of the distance from $(x,y,z)$ to $(ta,tb,tc)$. Note that $f(t)\ge 0$ for all $t$.
But by the standard distance formula, we have
$$f(t)=(ta-x)^2+(tb-y)^2+(tc-z)^2.\tag{$1$}$$
Expand. We find that
$$f(t)=t^2(a^2+b^2+c^2)-2t(ax+by+cz)+x^2+y^2+z^2.$$
The above quadratic is always $\ge 0$ precisely if the discriminant
$$4(ax+by+cz)^2-4(a^2+b^2+c^2)(x^2+y^2+z^2)$$
is $\le 0$. That gives the desired inequality.
We also can get the condition for equality out of this. The discriminant is $0$ precisely if $f(t)=0$ has a double root $t_0$. that is the case iff $x=at_0$, $y=bt_0$, and $z=ct_0$, that is, precisely if the vector $(x,y,z)$ is a multiple of $(a,b,c)$.
Remark:: We cheated, basically the same proof works in $\mathbb{R}^n$ for any $n$.  We do not even have to know the distance formula, since we can just define $f(t)$ as in Equation $(1)$.
A: Well...the cauchy-schwarz inequality states that $|\vec{a}\cdot \vec{b}|\leq ||\vec a|| ||\vec b||$.
Well, we know that $\vec{a}\cdot \vec{b} = ||\vec a||||\vec b||cos\theta$.
$cos\theta$ can be 1 at most and -1 at least, and so that means that  $|\vec{a}\cdot \vec{b}|\leq ||\vec a|| ||\vec b||$.
As simple as that =)
My solution assumes that you know that $\vec{a}\cdot \vec{b} = ||\vec a||||\vec b||cos\theta$.  If you need to prove that, that is a quick geometric proof where you graph any $\vec a$ and and any $\vec b$ and you make the vector $\vec a + \vec b$ and you use the law of cosines in order to get that expression.
