# Inconventional Integral inequality

$$\int_a^bw(x)|f(x)||g(x)|\;dx \le \left(\int_a^bw(x)\;dx\right) \max_{a\le x\le b}|f(x)|\cdot \max_{a\le x\le b}|g(x)|$$

I don't really understand this integral inequality. How do I go about conceptualising this

-
I changed your picture to LaTeX. Please confirm that I haven't messed something up, since your picture was slightly hard to read. –  apnorton Jul 19 '14 at 22:16
$w(x) \geqslant 0$, presumably, for otherwise the inequality could be wrong. You make the integrand larger by replacing $\lvert f(x)\rvert$ by its maximum. The same for $g$. Then you pull the constants out of the integral. –  Daniel Fischer Jul 19 '14 at 22:17

$\displaystyle{|f(x)| \leq \max_{a \leq x \leq b} |f(x)|}$ $\ \ \ \$ ($\displaystyle{\max_{a \leq x \leq b} |f(x)|}$ is a constant, since it the value of the function at a specific $x$)

$\displaystyle{|g(x)| \leq \max_{a \leq x \leq b} |g(x)|}$ $\ \ \ \$ ($\displaystyle{\max_{a \leq x \leq b} |g(x)|}$ is a constant, since it the value of the function at a specific $x$)

$$\int_a^b w(x) |f(x)| |g(x)| dx \leq \int_a^b w(x) \max_{a \leq x \leq b} |f(x)| \max_{a \leq x \leq b} |g(x)|dx= \\ \left ( \int_a^b w(x) dx \right ) \max_{a \leq x \leq b} |f(x)| \max_{a \leq x \leq b} |g(x)|dx$$

We can write the values $\displaystyle{\max_{a \leq x \leq b} |f(x)|}$ and $\displaystyle{\max_{a \leq x \leq b} |g(x)|}$ out of the integral, since they are constants and so independent from $x$.

-