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Is there an identity that says $|\sqrt {a^2+x^2} - \sqrt {a^2+y^2}| \leq |\sqrt {x^2} - \sqrt {y^2}|$?

Because of the nature of the square root function, its derivative monotonically decreases. so differences "further up" the function would be less than those lower down.

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Yes. $$ \left|\sqrt {a^2+x^2} - \sqrt {a^2+y^2}\right| =\frac{\lvert x^2-y^2\rvert}{\left|\sqrt {a^2+x^2} + \sqrt {a^2+y^2}\right|} = |\sqrt {x^2} - \sqrt {y^2}|\cdot \frac{|\sqrt {x^2} + \sqrt {y^2}|}{\left|\sqrt {a^2+x^2} + \sqrt {a^2+y^2}\right|} \leq |\sqrt {x^2} - \sqrt {y^2}| $$

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    $\begingroup$ This has made my life significantly easier :D. $\endgroup$ – user197848 Feb 11 '16 at 10:46
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This is true. You can see this by assuming $x>y$ without losing generality and then differentiating \begin{equation} f(a)=\sqrt{a^2+x^2}-\sqrt{a^2+y^2} \end{equation} with respect to $a$. Derivative is negative hence $f$ is decreasing function of $a$ and is maximized at $0$.

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