Prove that $$a+b+c \geq \sqrt{b^2+c^2-a^2} + \sqrt{c^2+a^2-b^2} + \sqrt{a^2+b^2-c^2}$$ where $a, b, c$ are sidelengths of a non-obtuse triangle. $$ $$ I found this inequality in an old Crux Mathematicorum journal but no solution was provided. I used Cauchy-Schwarz to handle the square roots: $$ \sqrt{(1+1+1)\left((b^2+c^2-a^2)+(c^2+a^2-b^2)+(a^2+b^2-c^2)\right)} \geq \sqrt{b^2+c^2-a^2} + \sqrt{c^2+a^2-b^2} + \sqrt{a^2+b^2-c^2}$$
but I don't know how to proceed from here as the inequality $a+b+c \geq \sqrt{3(a^2+b^2+c^2)}$ is simply false.
Any hints or solutions would be appreciated. Thanks
Because by C-S $$\sum_{cyc}a=\frac{1}{2}\sum_{sys}\sqrt{(1+1)(a^2+b^2-c^2+a^2+c^2-b^2)}\geq$$ $$\geq\frac{1}{2}\sum_{cyc}\left(\sqrt{a^2+b^2-c^2}+\sqrt{a^2+c^2-b^2}\right)=\sum_{cyc}\sqrt{a^2+b^2-c^2}.$$