Prove the inequality $\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$ Let $a,b,c -$ triangle side and $a+b+c=1$. Prove the inequality
$$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$$
My work so far:
1) $a^2+b^2=c^2-2ab\cos \gamma \ge c^2-2ab$
2) $$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}\le3\sqrt{\frac{2(a^2+b^2+c^2)}3}$$

A similar approach to the one suggested by Jack to obtain the loose inequality: the function $\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}$ is convex in $a,b,c$ being the sum of three convex functions (they're the restriction of the Euclidean norm to the planes $\{a=0\}$ etc.). The domain of the function is
$$\{a,b,c\geq 0\}\cap\{a+b+c=1\}\cap \{a\leq b+c\}\cap\{b\leq c+a\}\cap\{c\leq a+b\}$$
which is a polygon in $\mathbb{R}^3$, actually an equilateral triangle inside the plane $\{a+b+c=1\}$ if you want to visualize it, with vertices $(0,\frac12,\frac12)$ and cyclicals. Therefore to know the maximum it is sufficient to look at the extremal points of the domain, which are the three vertices. 
In this way you obtain the loose inequality. The strict one comes from the fact that you want nondegenerate triangles, and you might have to work a bit to prove that restricted to the sides of the triangle the function is strictly convex.
