Inequality involving lengths and triangles I was quite sure this would have been asked before but I couldn't find it, so here goes:
If $\displaystyle BC<AC<AB \hspace{5pt} (\alpha<\beta<\gamma)$, show $\displaystyle PA+PB+PC<\beta+\gamma \hspace{3pt}$ (where $P$ arbitrary in the interior of the triangle).

I haven't actually been able to make a simple yet convincing attempt (although I have got a messy solution). I know that there is a nice and elegant way to do it, but it isn't coming to me.

The Fermat-Torricelli point of a triangle $ABC$ is the point $P$ such that $PA+PB+PC$ is minimized. It is usually found through the following lines: Assume, for first, that $ABC$ has no angle greater than $120^\circ$ and take a point $Q$ inside $ABC$. Let $Q'$ the image of $Q$ under a rotation of $60^\circ$ around $B$, $A'$ likewise:

Then $Q'A'=QA$ and $BQ=BQ'=QQ'$, hence:
$$ QA+QB+QC = A'Q'+Q'Q+QC \leq CA', $$
and the Fermat-Torricelli point lies in the intersection of the lines through a vertex of $ABC$ and the corresponding vertex of a equilateral triangle built on the opposite side, externally to $ABC$.
However, if $ABC$ has an angle $\geq 120^\circ$, the Fermat-Torricelli point is just the opposite vertex to the longest side.
Can you finish from there?

Update: I also have a one-shot-one-kill solution. The function that maps $P$ to $PA+PB+PC$ is a convex function, as a sum of convex functions. It follows that $PA+PB+PC$ attains its maximum in a vertex of $ABC$, since $ABC$ is a convex set. If $BC<AC<AB$, such a maximum is obviously attained in $A$, and the claim follows.
