# How to prove this inequality (like triangle inequality)

[Question] $$\left|\sqrt {x^2+y^2}-\sqrt {x^2+z^2}\right| \le |y-z|$$

[My Effort] \begin{align} &I_1=\sqrt {x^2+y^2} \le |x|+|y|\\ &I_2=\sqrt {x^2+z^2} \ge \left||x|-|z|\right|\\ \implies\\ &I_1-I_2 \le |x|+|y|-\left||x|-|z|\right|=\begin{cases} |y|+|z|, &\mbox {if }(|x|\ge|z|)\\ 2|x|+|y|-|z|,&\mbox {if }(|x|<|z|) \end{cases} \end{align} Either way, I can't reduce to $|y-z|$.

This is exactly the (so called inverse) triangle inequality in $\mathbb R^2$ applied to the vectors $(x,y)$ and $(x,z)$: $$\left|\sqrt {x^2+y^2}-\sqrt {x^2+z^2}\right|=\left|\|(x,y)\|_2-\|(x,z)\|_2\right|\leq \|(x,y)-(x,z)\|_2\\ =\|(0,y-z)\|_2=\sqrt{0^2+(y-z)^2}=|y-z|$$
Hint: The result is obvious if $x=0$. If $x\ne 0$, multiply top and bottom on the left by $\sqrt{x^2+y^2}+\sqrt{x^2+z^2}$ ("rationalize" the numerator). Then use the fact that $\frac{|y+z|}{\sqrt{x^2+y^2}+\sqrt{x^2+z^2}}\le 1$.
• I had a typo in an earlier comment. Using my suggestion, the new top is $|y-z||y+z|$. And we have $|y+z|\le |y|+|z|$, while $\sqrt{x^2+y^2}+\sqrt{x^2+z^2}\ge |y|+|z|$. Commented Oct 9, 2015 at 15:31