# Proving Cauchy Inequality from Hölder's inequality?

I wish to prove $$\left(\sum_{i=1}^n a_ib_i\right)^2 \leq \left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)$$ for really numbers $a_1,\dots,a_n,b_1,\dots,b_n$, by using Hölder inequality.

My plan was to define random variables $X = a_i$ with probability $\frac1n$ and $Y=b_i$ with probability $\frac1n$. Then, using Hölder's, I can get

$$E\vert X\cdot Y\vert \leq \left[\left(\frac{1}{n}\sum_{i=1}^n a_i^2\right)\left(\frac{1}{n}\sum_{i=1}^n b_i^2\right)\right]^{1/2}$$ and by squaring that I can get the RHS correct.

However, the LHS is not correct with this definition of $X,Y$. Specifically, $X\cdot Y = a_ib_j$ with probability $\frac{1}{n^2}$ so $$E(X\cdot Y) =\frac{1}{n^2}\sum_{i=1}^na_i\sum_{j=1}^n b_j$$ which is where I'm stuck. Also, I don't quite see what $E\vert XY\vert$ would be since $a_i b_j$ can be negative, so the absolute value matters.

If $XY$ was uniformly distributed over $\{a_ib_i\}$ then I would be golden.

Hints are much appreciated.

• To prove C-S using probability theory you have check that any theorems from probability that are used were not derived using C-S at any stage, which means thoroughly checking their proofs. Are you familiar with the elementary algebraic proof of C-S? – DanielWainfleet Jun 21 '17 at 5:52

In the standard formulation of Holder's inequality, the $p=q=2$ case is just the Cauchy-Schwarz inequality.

For your probabilistic version, you need to align your variables $X$ and $Y$; throw a fair $n$-sided die, and let $X=a_i$ and $Y=b_i$ when it comes up $i$.

• I see. I was thinking about having $X,Y$ being dependent, but could not visualize how to make it happen. With the roll of the dice I understand. Thank you. I will accept once the required time passes. – user106860 Jun 21 '17 at 5:05

Holder it's the following thing.

Let $a_1$, $a_2$,..., $a_n$, $b_1$, $b_2$,..., $b_n$, $\alpha$ and $\beta$ be positive numbers. Prove that: $$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}\geq$$$$\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}$$

Thus, by Holder for $\alpha=\beta=1$ we obtain: $$\sum_{i=1}^na_i^2\sum_{i=1}^nb_i^2\geq\left(\sum_{i=1}^n\sqrt{a_i^2b_i^2}\right)^2=\left(\sum_{i=1}^n|a_ib_i|\right)^2\geq\left(\sum_{i=1}^na_ib_i\right)^2.$$ Done!

• I found the algebra manipulation going from the double sum to the right interesting, thanks. – user106860 Jun 21 '17 at 5:32