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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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    Well, aren't integrals defined in terms of sums?2012-08-18
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    Try to use the fact that $\int (a-cb)^2 \geq 0$ for all real numbers $c$ and then choose a clever value for $c$.2012-08-18
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    It's Cauchy--[Schwarz](http://en.wikipedia.org/wiki/Hermann_Amandus_Schwarz) not [Schwartz](http://en.wikipedia.org/wiki/Laurent_Schwartz). I fixed it for you.2012-08-18
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    @NateEldredge: Good catch, Thanks!2012-08-18
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    @DilipSarwate: I see, clever.2012-08-18
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    @PeterTamaroff : yes, taht's what I ended up doing.2012-08-18
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    @user1311390 Yes, clever. Try it with $$c = \frac{\int ab}{\int b^2}$$ (assuming that $\int b^2 > 0$), and work out a separate proof for the case when $\int b^2 = 0$.2012-08-18

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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)2012-08-18