The Cauchy-Schwarz integral inequality is as follows:
$$
\displaystyle \left({\int_a^b f \left({t}\right) g \left({t}\right) \ \mathrm d t}\right)^2 \le \int_a^b \left({f \left({t}\right)}\right)^2 \mathrm d t \int_a^b \left({g \left({t}\right)}\right)^2 \mathrm d t
$$
How do I prove this using multivariable calculus methods, preferably with double integrals?
Best Answer
If by double-integral you mean the identity: $$\frac{1}{2}\int_a^b\int_a^b (f(x)g(y) - g(x)f(y))^2\,dx\,dy \\= \int_a^b f^2(x)\,dx\int_a^b g^2(x)\,dx - \left(\int_a^b f(x)g(x)\,dx\right)^2$$
Then note that the integrand $\displaystyle (f(x)g(y) - g(x)f(y))^2 \ge 0$, hence the inequality follows.