$|{\sqrt{a^2+b^2}-\sqrt{a^2+c^2}}|\le|b-c|$ where $a,b,c\in\mathbb{R}$ Show that $|{\sqrt{a^2+b^2}-\sqrt{a^2+c^2}}|\le|b-c|$ where $a,b,c\in\mathbb{R}$
I'd like to get an hint on how to get started. What I thought to do so far is dividing to cases to get rid of the absolute value. $(++, +-, -+, --)$
but it looks messy. I'm wondering if there is any nicer way to solve it.
Would love to hear some ideas.
Thanks in advance!
 A: It will be easy if you think $\sqrt{a^2 + b^2}$ as the euclidean distance. Consider the three points $A(a, 0), B(0, b), C(0, c)$. Then the inequality can be transformed into the triangular inequality $\lvert \overline{AB} - \overline{AC} \rvert \le \overline{BC}$
A: Use the formula: $${\sqrt{x}-\sqrt{y}}= {x-y\over \sqrt{x}+\sqrt{y}}$$
in your case you get:
$${\sqrt{a^2+b^2}-\sqrt{a^2+c^2}}= {a^2+b^2-(a^2+c^2)\over \sqrt{a^2+b^2}+\sqrt{a^2+c^2}}$$
so you have to prove (if you assume $b\geq c$, which you can): $$b+c\leq \sqrt{a^2+b^2}+\sqrt{a^2+c^2}$$
and this is trivaly...
A: For all $a,b,c \in \mathbb{R}$,
\begin{align}
0 &\leq (b-c)^2\\
2bc &\leq b^2 + c^2\\
2a^2bc &\leq a^2b^2 + a^2c^2\\
a^4 + 2a^2bc + b^2c^2 &\leq a^4 + a^2b^2 + a^2c^2 + b^2c^2\\
a^2 + bc&\leq \sqrt{a^4 + a^2b^2 + a^2c^2 + b^2c^2} \label{1}\tag{1}\\
2a^2 + 2bc&\leq 2\sqrt{a^4 + a^2b^2 + a^2c^2 + b^2c^2}\\
2a^2 -2\sqrt{a^4 + a^2b^2 + a^2c^2 + b^2c^2}&\leq -2bc\\
2a^2 +b^2 + c^2 -2\sqrt{a^4 + a^2b^2 + a^2c^2 + b^2c^2}&\leq b^2 -2bc + c^2\\
(a^2 +b^2) -2\sqrt{(a^2+b^2)(a^2+c^2)} + (a^2 + c^2)&\leq b^2 -2bc + c^2\\
\end{align}
and then take the square root of both sides.
Note that it was not necessary to use absolute values the first time we took a square root $(\ref{1})$ because the right hand side is understood to be positive.

