Prove if$~\vert x \vert < z~$ and $~\vert y \vert < z~$ then $~\Big\vert \dfrac{x+y}{2}\Big\vert + \Big\vert \dfrac{x-y}{2}\Big\vert < z~$

Question

Let $x~$,$y~$,$z~\in\mathbb{R}$
Prove that : if$~\vert x \vert < z~$ and $~\vert y \vert < z~$ then $~\Big\vert \dfrac{x+y}{2}\Big\vert + \Big\vert \dfrac{x-y}{2}\Big\vert < z~$
I know how to prove it by separating cases, but I need a Hint for a direct proof.

Answer 1

Without loss of generality you can assume that $x^2 \ge y^2$. Then $$ \Bigl(|x+y| + |x-y|\Bigr)^2 = (x+y)^2 + (x-y)^2 + 2|(x+y)(x-y)| = 4x^2 < 4 z^2. $$

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More precisely we have, using that $\max(a, b) = \frac 12 (a+b+|a-b|$: $$ \Bigl(|x+y| + |x-y|\Bigr)^2 = (x+y)^2 + (x-y)^2 + 2|(x+y)(x-y)| \\ = 2x^2 + 2y^2 + 2|x^2-y^2| = 4\max(x^2, y^2) $$ and therefore $$ \left\vert \frac{x+y}{2}\right\vert + \left\vert \frac{x-y}{2}\right\vert = \max(|x|, |y|) \, . $$

Answer 2

Notice that for $x\rightarrow -x$ we get the same inequality.Symmetrically it follows for $y$.Now WLOG $x\ge y$,:: $x,y\ge 0$ Hence $$|\frac{x+y}{2}|+|\frac{x-y}{2}|=\frac{x+y}{2}+\frac{x-y}{2}=x<z $$