I'm taking an online course in cryptography and trying to follow one of the proofs. [ For context, the proof is here> https://youtu.be/4vbwxPR_D2U?t=449 ].
There seems to be an assumption (which is not proven by the lecturer) that, given $a,b,r,y\geq 0$
If we have $$ | a -r | \leq y \quad \text{and}\quad | b -r | \leq y $$ the above implies that $$ |a - b| \leq 2y $$ I tried proving this to myself, but came up dry. Hopefully my distillation of the professor's assumption as stated above is correct and provable. If it is provable and someone can sketch the proof out, I'd be super grateful.
This is just a straightforward application of the triangle inequality: $$\begin{align*} |a-b| &= |a-r-b+r|\\ &\le |a-r| + |r-b|\\ &= |a-r| + |b-r|\\ &\le y + y \\ &= 2y \end{align*}$$ In general, $$ |x+y| \le |x|+|y| $$ for complex numbers $x,y$. This can be generalized to vector norms and to metric spaces in general as well.