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There is a problem in my book that says:

Show that (assuming $f,g,f^2, g^2, $and$ fg$ are integrable) $$\int_{a}^{b}fg\leq \sqrt{\int_{a}^{b}f^2}\sqrt{\int_{a}^{b}g^2}$$. I know this is the Cauchy-Shwarz Inequality for integrals, but my book says to prove it by defining $$p(\lambda)=\int_{a}^{b}[f-\lambda g]^2$$ for each real number $\lambda$, and then show that $p(\lambda)$ is a quadratic polynomial for which $p(\lambda)\geq0$, for all $\lambda$, and thus the discriminant is not positive. How does this help?

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You can put an absolute value sign on the left side for a stronger statement. –  NotNotLogical Apr 23 at 2:23

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up vote 4 down vote accepted

Since $p(\lambda)$ is the integral of a square, it is non-negative: $$ \begin{align} 0 &\le p(\lambda)\\ &=\int f(x)^2\,\mathrm{d}x-2\lambda\int f(x)g(x)\,\mathrm{d}x+\lambda^2\int g(x)^2\,\mathrm{d}x\\ \end{align} $$ Therefore, for $\lambda\gt0$, $$ \int f(x)g(x)\,\mathrm{d}x \le\frac1{2\lambda}\int f(x)^2\,\mathrm{d}x+\frac\lambda2\int g(x)^2\,\mathrm{d}x $$ If we set $\lambda=\sqrt{\frac{\int f(x)^2\,\mathrm{d}x}{\int g(x)^2\,\mathrm{d}x}}$, we get $$ \int f(x)g(x)\,\mathrm{d}x \le\sqrt{\int f(x)^2\,\mathrm{d}x}\sqrt{\int g(x)^2\,\mathrm{d}x} $$

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$$p(\lambda)=\lambda^2\int g^2 -2\lambda \int fg +\int f^2$$

Now what is $b^2-4ac$, in terms of the question you mention? Cancel $\lambda^2$ from that being $\geq 0$...

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Thank you I understand now –  Paul Malinowski Apr 23 at 2:24

This is basically the proof of the Cuachy-Shwarz inequality. Look it up and you are good to go.

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