= Solution
Assume real square-integrable functions on the same integration domain. Put $A=\int f^2$, $B=\int g^2$, and $C=\int fg$. If $B=0$, then $g=0$ almost everywhere, so $C=0$ and the claim follows. If $B>0$, nonnegativity of a square gives
$$
0\leq\int\left(f-\frac CBg\right)^2\,dx
=A-\frac{C^2}{B}.
$$
Multiplying by $B$ proves the <integral> form of the <Cauchy-Schwarz inequality>:
$$
\boxed{\left(\int f^2\,dx\right)\left(\int g^2\,dx\right)
\geq\left(\int fg\,dx\right)^2}.
$$
For nonzero functions, equality holds exactly when they are proportional almost everywhere. The zero-function cases also give equality.
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