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

$\displaystyle{|f(x)| \leq \max_{a \leq x \leq b} |f(x)|}$ $ \ \ \ \ $ ($\displaystyle{\max_{a \leq x \leq b} |f(x)|}$ 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 $x$.
