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[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 $|y-z|$.

Thanks in advance!

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2 Answers 2

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This is exactly the (so called inverse) triangle inequality in $\mathbb R^2$ applied to the vectors $(x,y)$ and $(x,z)$: $$\left|\sqrt {x^2+y^2}-\sqrt {x^2+z^2}\right|=\left|\|(x,y)\|_2-\|(x,z)\|_2\right|\leq \|(x,y)-(x,z)\|_2\\ =\|(0,y-z)\|_2=\sqrt{0^2+(y-z)^2}=|y-z|$$

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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 $\frac{|y+z|}{\sqrt{x^2+y^2}+\sqrt{x^2+z^2}}\le 1$.

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  • $\begingroup$ Sorry, my English is not very good. "top and bottom", what does that mean? $\endgroup$
    – Winter Young
    Commented Oct 9, 2015 at 15:14
  • $\begingroup$ Oh, I saw "multiply" as "multiple"... I see what you mean. I tried that, no luck. Mind some details? $\endgroup$
    – Winter Young
    Commented Oct 9, 2015 at 15:18
  • $\begingroup$ 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 $\sqrt{x^2+y^2}+\sqrt{x^2+z^2}\ge |y|+|z|$. $\endgroup$
    – André Nicolas
    Commented Oct 9, 2015 at 15:31
  • $\begingroup$ You are welcome. The answer by Svetoslav is less computational, and I think better. $\endgroup$
    – André Nicolas
    Commented Oct 9, 2015 at 16:08

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