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.
How to prove Cauchy-Schwarz Inequality in $R^3$?
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0Since you have proved it for $\mathbb{R}^2$, try using induction. – 2012-09-12
4 Answers
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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0@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). – 2012-09-16
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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0@PeterTamaroff: Thanks for the edits. I hope the above clarifies... – 2012-09-13
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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0@RobertMiller: Thanks for catching it! Fixed. – 2012-09-13
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.