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.

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Since you have proved it for $\mathbb{R}^2$, try using induction. –  tcmtan Sep 12 '12 at 23:17

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 $x^2\geq 0$.

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Peter, I appreciate your response and all the other responses. I have selected this answer as it is the closest to what I had already started. I was missing the 2 at the start. Thanks again for clearly showing the steps. –  RexBanner Sep 13 '12 at 21:07
@RexBanner Anytime. –  Pedro Tamaroff Sep 13 '12 at 22:05
@Peter Tamaroff: Does Cauchy-Schwarz state that $x_1y_1+x_2y_2+x_3y_3\leq \sqrt{x_1^2+x_2^2+x_3^3}\sqrt{y_1^2+y_2^2+y_3^3}$? And I don't think that your first example is a good one for what Cauchy-Schwarz inequality is. (Chris). –  Chris's sis Sep 16 '12 at 11:54

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.

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It would be good if you could explain the indices notation. –  Pedro Tamaroff Sep 13 '12 at 1:45
@PeterTamaroff: Thanks for the edits. I hope the above clarifies... –  copper.hat Sep 13 '12 at 1:50

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.

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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 $(1)$.

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Condition on discriminant should be less than or equal to zero? –  Robert Miller Sep 13 '12 at 0:59
@RobertMiller: Thanks for catching it! Fixed. –  André Nicolas Sep 13 '12 at 1:09