Prove that $\max(a+x, b+x) = \max(a,b) + x$

Question

I want to prove the following simple identity: $$\max(a+x, b+x) = \max(a,b) + x$$

Answer 1

We look at cases:

Case 1: $a < b$. Then we have $\max(a+x, b+x) = b + x$, $\max(a,b) + x = b + x$.

Case 2: $a \geq b$. Then we have $\max(a + x, b + x) = a + x$, $\max(a,b) + x = a + x$.

So the two formulas are equivalent in all cases.

Answer 2

We can use $\max(a, b) =\dfrac{a+b+|a-b|}{2} $.

Then

$\begin{array}\\ \max(a+x, b+x) &=\dfrac{(a+x)+(b+x)+|(a+x)-(b+x)|}{2}\\ &=\dfrac{a+b+2x+|a-b|}{2}\\ &=\dfrac{a+b+|a-b|}{2}+x\\ &=\max(a, b)+x\\ \end{array} $

Answer 3

It's evident that \begin{align*} a& \leqslant \max\{a+x,b+x\}-x,\\ b& \leqslant \max\{a+x,b+x\}-x. \end{align*} Thus $$\max\{a,b\}\leqslant \max\{a+x,b+x\}-x.\tag{1}$$ On the other hand, \begin{align*} a+x& \leqslant \max\{a,b\}+x,\\ b+x& \leqslant \max\{a,b\}+x. \end{align*} Hence $$\max\{a+x,b+x\}\leqslant \max\{a,b\}+x.\tag{2}$$ (1) together with (2) shows that $$\max\{a+x,b+x\}=\max\{a,b\}+x.$$