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Proving the discrete version of Cauchy-Schwarz is easy:

$$ \left(\sum_{i} a_i^2\right) \left(\sum_i b_i^2\right) \geq \left(\sum_i a_ib_i\right)^2 $$

can be done via the determinant of the quadratic formula.

Now, however, I want to prove the continuous version, which states:

$$ \int a^2 \int b^2 \geq \left(\int ab\right)^2$$

How do I prove this?

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2  
Well, aren't integrals defined in terms of sums? –  Pedro Tamaroff Aug 18 '12 at 4:01
3  
Try to use the fact that $\int (a-cb)^2 \geq 0$ for all real numbers $c$ and then choose a clever value for $c$. –  Dilip Sarwate Aug 18 '12 at 4:03
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It's Cauchy--Schwarz not Schwartz. I fixed it for you. –  Nate Eldredge Aug 18 '12 at 4:07
    
@NateEldredge: Good catch, Thanks! –  user36739 Aug 18 '12 at 4:10
    
@DilipSarwate: I see, clever. –  user36739 Aug 18 '12 at 4:11

1 Answer 1

up vote 3 down vote accepted

First note that $ab \leq \frac{a^2}{2} + \frac{b^2}{2}$. Then take $a = \frac{f}{(\int{f^2})^{\frac{1}{2}}}, b = \frac{g}{(\int{g^2})^{\frac{1}{2}}}$.

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I like this solution. It's much better than what I ended up with (expressing the integrals as reinmen sums, and arguing that the approximations are within epsilon) –  user36739 Aug 18 '12 at 4:09

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