Show Riemann sums satisfy $|f(b) - f(a)| * ||P||$

Question

Suppose $f$ is monotonically decreasing on interval [a,b]. For any partition $P = (a = x_0, x_1, ..., x_n = b)$ of [a,b] let $P_L$ denote the tagged partition with tags chosen to be the left endpoints and $P_R$ the tagged partition with tags chose to be the right endpoints of the corresponding intervals $I_j = [x_j-1, x_j], j = 1,2,...,n$.

Show:

$|S(f;P_L) - S(f;P_R)| \leq |f(b) - f(a)| * ||P||$

I definitely understand what Riemann sums are and what the question is asking but I don't know how to show this arithmetically without pointing to a graph. I tried playing with the summations a little but couldn't get anything. Any direction is appreciated.

In case there are confusions with the notation,

$S(f;P_L) = \sum\limits_{j=0}^{n-1} f(x_j^L)(x_{j+1} - x_j)$

$||P|| = max(x_{j+1} - x_j)$, ie, the largest width of a rectangle.

Answer 1

A hint:

Since $f$ is decreasing each subinterval $I_j:=[x_{j-1},x_j]$ contributes $$\bigl(f(x_{j-1})-f(x_j)\bigr)(x_j-x_{j-1})$$ to the difference $S(f,P_L)-S(f,P_R)$. Note that there is no need to bring absolute values into the balance sheet. Now sum it all up, using $x_j-x_{j-1}\leq\|P\|$ for all $j\in[n]$.