Then use the "vector space" proof when the vector space in question is a function space!
The "vector space" proof looks at the inner product of \vec{u}- \lambda\vec{v} with itself, getting |\vec{u}-\lambda\vec{v}|= <\vec{u}- \lambda\vec{v},\vec{u}= \lambda\vec{v}>= <\vec{u},\vec{u}>- 2\lambda<\vec{u},\vec{v}>+ \lambda^2<\vec{v},\vec{v}>\ge 0, for all \lambda because it is a "square".
In particular, if we let let
\lambda= \frac{<\vec{u},\vec{v}>}{<\vec{v},\vec{v}>}
that becomes
<\vec{u},\vec{u}>-2\frac{\vec{u},\vec{v}>^2}{<\vec{v},\vec{v}}+ \frac{\vec{u},\vec{v}>^2}{\vec{v},\vec{v}}
= <\vec{u},\vec{u}>- \frac{<\vec{u},\vec{v}>^2}{<\vec{v},\vec{v}>}\ge 0
so
<\vec{u},\vec{u}>\ge \frac{<\vec{u},\vec{v}>^2}{<\vec{v},\vec{v}>}
and
<\vec{u},\vec{u}><\vec{v},\vec{v}>\ge<\vec{u},\vec{v}>^2
That is precisely the proof for "vector spaces" at the site I linked to. Now, the set of all integrable functions IS an inner product space with innerproduct given by <f, g>= \int fg.
Replacing <\vec{u},\vec{v}> by \int fg , <\vec{u}, \vec{v}> with \int f^2, and <\vec{v}, \vec{v}> with \int g^2 gives exactly the inequality you post. That's the whole point of "abstraction" in mathematics- you don't have to prove a lot of different versions of the same thing.
