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: Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, $x$, $y$ and $z$ are positives and we need to prove that:
$$\sum\limits_{cyc}\sqrt{2x^2+y^2+z^2+2xy+2xz}\leq(2+\sqrt2)(x+y+z)$$
By C-S $$\left(\sum\limits_{cyc}\sqrt{2x^2+y^2+z^2+2xy+2xz}\right)^2\leq\sum\limits_{cyc}\frac{2x^2+y^2+z^2+2xy+2xz}{\sqrt2x+y+z}\sum\limits_{cyc}(\sqrt2x+y+z)$$
Thus, it remains to prove that
$$\sum\limits_{cyc}\frac{2x^2+y^2+z^2+2xy+2xz}{\sqrt2x+y+z}\leq(2+\sqrt2)(x+y+z)$$ or
$$\sum\limits_{sym}((1+\sqrt2)x^3y+2.5x^2y^2+(3\sqrt2+1.5)x^2yz)\geq0$$
which is obvious. Done!
A: Just a sketch of a possible proof (a variational, mixing variables approach). Assume that $a\leq b\leq c$ and show that if you replace $a$ with $a-\varepsilon$ and $b$ with $b+\varepsilon$, the convexity of $f(x)=\sqrt{1+x^2}$ grants that the LHS increases (after such substitution, the order of $b$ and $c$ may change. That is not an issue, it actually helps us to be sure we always deal with side lengths of a triangle). It follows that the supremum of the LHS is achieved when $a=0$. After that, the problem is trivial.
A: 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.
A: The triangle condition is equivalent to $0<a<1/2$, $0<b<1/2$ and $0<c<1/2$. Assume $c=\min(a, b, c)$, then 
$$\sqrt{a^2+b^2} + \sqrt{b^2+c^2} + \sqrt{a^2+c^2} \le \sqrt{a^2+b^2} + \sqrt{b^2+bc} + \sqrt{a^2+ac} < \sqrt{a^2 + b^2} + (b+c/2) + (a+c/2) = \sqrt{a^2+b^2} + 1 < \sqrt{(1/2)^2+(1/2)^2} + 1 = 1+\frac{\sqrt{2}}{2}$$
A: Try to use Lagrange multipliers to solve this problem.
max $[\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}+\lambda(a+b+c-1)]$
