I am wondering if someone can provide some geometric intuition, or some simple way to visualize why $$ c-d<a-b \implies b<a+d-c $$
The way I have been trying to do this is to think of $a,b,c,d$ as points in $\mathbb{R}$ and $a-b, c-d$ as the space between them. but
- $c-d<a-b$ doesn't necessarily mean that $\vert c-d\vert < \vert a-b\vert$,
- and when rearranged to be $b<a+d-c$, the left hand side is no longer a "space between two points"
So my approach didn't really help me.
First, consider two real numbers $a,b$ such that $a<b$. Picture a number line: this means $a$ is to the left of $b$. If we add a real number $c$ to both sides, then we have $a+c<b+c$. The geometric intuition here should be clear: we have shifted the point $a$ either right or left by a distance $|c|$ to the point $a+c$, and $b$ to $b+c$; since we shifted these points together, we still have $a+c$ to the left of $b+c$.
For your particular inequality, we just have to take this idea and apply it a couple times: start with $c-d<a-b$; add $b$ to both sides; then add $d-c$. Geometrically, the points $c-d$ and $a-b$ are first shifted either right or left by a distance $|b|$, then shifted right or left by a distance $|d-c|$.