Show that $|\int^b_a{f(x)g(x)dx}| ≤$ ($\int^b_a{f^2(x)dx})^{1/2}$ . $(\int^b_a{g^2(x)dx})^{1/2}$ 
Let $f,g : [a,b] \rightarrow \mathbb{R}$ be Riemann integrable. Show
  that    $|\int^b_a{f(x)g(x)dx}| ≤$ ($\int^b_a{f^2(x)dx})^{1/2}$ .
  $(\int^b_a{g^2(x)dx})^{1/2}$

The proof reminded me of the Cauchy-Schwartz proof for Real numbers. I investigated   

$k(t) = \int^b_a{(tf(x)+g(x))^{1/2}}dx$

to see that with this polynomial, the sign is always positive and hopefully see how the inequality can come out. Any ideas on the proof?
 A: It is true this is a Cauchy Schwarz inequality: consider the inner product 
$$\langle f,g\rangle =\int_a^b fg$$
and copy the proof of Cauchy Schwarz you're thinking about. Alternatively, and in the lines of the more general Hölder inequalities in $L^p$ spaces, you can do the following. For every $a,b>0$, it is true that $$ab\leqslant \frac{a^2+b^2}2$$
Assume first that $f,g$ are such that $\int_a^b f^2=\int_a^b g^2=1$. Then integrating the inequality $$|fg| \leqslant \frac{f^2+g^2}2$$
gives $$\int_a^b |fg|\leqslant 1$$
You may now deduce the general case by normalizing any given $f,g$ (provided they are nonzero), that is, setting $f' = (\int_a^b f^2)^{-1} f$,$g'= (\int_a^b g^2)^{-1} g$ and using the above. 
A: Using Cauchy-Schwartz for finite sums, we get
$$|\sum_{k=1}^{n}f(x_k)g(x_k)\Delta x_k| = |\sum_{k=1}^{n}f(x_k)(\Delta x_k)^{1/2}g(x_k)(\Delta x_k)^{1/2}|$$ $$ \le \left(\sum_{k=1}^{n}f(x_k)^2\Delta x_k \right)^{1/2}\cdot \left (\sum_{k=1}^{n}g(x_k)^2\Delta x_k \right )^{1/2}.$$
Now take the limit as the mesh size of the partition $\to 0$ to get the desired result.
