Prove and give geometric meaning of $|\sqrt{a^2+b^2} - \sqrt{a^2+c^2} | < |b-c| $ 
Carefully, prove that
$$ | \sqrt{a^2+b^2} - \sqrt{a^2+c^2} | \leq |b-c| $$
and give a geometric interpretation.

pf
We observe that $| \sqrt{a^2+b^2} - \sqrt{a^2+c^2} | = \dfrac{ |b^2 - c^2 | }{\sqrt{a^2+b^2} + \sqrt{b^2+c^2} } \leq \dfrac{ |b^2-c^2| }{\sqrt{b^2} + \sqrt{c^2} } = \dfrac{ |b+c | }{|b| + |c| } \cdot |b-c|  $
Since by the triangle ineqality $|b+c| \leq |b| + |c| \implies \dfrac{ |b+c| }{|b| + |c| } \leq 1$, then the required result is proved!
Now, as for the hardest part, I dont see a geometric interpretation of this inequality. Can someone enlighten me?

Let $B(a,b)$ and $C(a,c)$ be the two vecrors. Then,
$$| |B|-|C||<| B-C| = |BC|$$
which translates To
$$| \sqrt{a^2+b^2} - \sqrt{a^2+c^2} | \leq |b-c| $$
Geometrically, the above inequality means the length difference of two triangle sides is less than the length of the third side.
