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?

As pointed out in the comments, the points $(0,0), (a,b), (a,c)$ form a triangle. Using the triangle inequality,
$$\sqrt{a^2 + b^2} + |b-c| \gt \sqrt{a^2 + c^2}$$
and
$$\sqrt{a^2 +c^2} + |b-c| \gt \sqrt{a^2 +b^2}$$
$$\implies \left|\sqrt{a^2+b^2}-\sqrt{a^2+c^2} \right| \lt |b-c|$$
