# What's the reason for this step in proof of reversed triangle inequality?

In proof of reversed triangle inequality $|x - y| \geq ||x| - |y||$, which is derived from triangle equality $|x + y| \leq |x| + |y|$ there is a step where triangle inequality is transformed into $$|(x-y)+ y| = |x| \leq |y| + |x - y|.$$

What is the logic behind that transformation? If I understood that correctly, we put $x = x - y$ in this case, to make $|x|$ term on LHS. This is also permissible to do for inequality, as inequality will remain the same.

Do I understand it correctly, or is there something else I do miss/do not understand?

• Another method is that since both sides are non-negative, we have $|x-y|\geq |\;|x|=|y|\;|\iff$ $|x-y|^2\geq |\;|x|-|y|\;|^2\iff$ $x^2-2xy+y^2\geq |x|^2-2|x|\cdot |y|+|y|^2\iff$ $|x|^2-2xy+|y|^2\geq |x|^2-2|x|\cdot |y|+|y|^2\iff$ $-2xy\geq 2|x|\cdot |y|\iff$ $xy\leq |x|\cdot |y|=|xy|.$ The disadvantage of this method is that it can't be used in more general contexts such as normed linear spaces. – DanielWainfleet Jun 7 '17 at 6:38

$$|x| = |x|$$

$$|x+0| = |x|$$

$$|(x+y)-y| = |x|$$

There is no change of variables in that step; adding a $0$ to only one side is a viable way to prove many different theorems, such as this one.

• And adding $-y$ to $x$ in RHS is permissible too, following this logic? So, basically, it won't change inequality, right? – Accelerate to the Infinity Oct 10 '16 at 18:04

Basically, start with |x| = |x + 0|. But 0 = y - y, so |x + 0| = |x + y - y| = |(x-y) + y|.

By the triangle inequality, |(x-y) + y| $\le$ |x-y| + |y|.

Think about it as an addition of 0 to x, rather than a substitution.

Is that what you were asking?

In fact, you can write this proof by changing variables (it's not the most common way, but if it helps, no problem): \begin{align*} X &= x - y \\ Y &= y \end{align*} Now TI with $X$ and $Y$ gives: \begin{align*} |X + Y| &\leqslant |X| + |Y| \\ |(x-y)+y | &\leqslant |x-y| + |y|\\ |x| &\leqslant |x-y| + |y|\\ \end{align*} And after that : $$|x| - |y| \leqslant |x-y|$$

Etc.