If $x+y+z=1$ prove $ \sqrt{x+\frac{(y-z)^{2}}{12}}+\sqrt{y+\frac{(z-x)^{2}}{12}}+\sqrt{z+\frac{(x-y)^{2}}{12}} \leq \sqrt{3} $ Question -
Let $x, y, z$ be non-negative real numbers which satisfies $x+y+z=1$. Prove that
$$
\sqrt{x+\frac{(y-z)^{2}}{12}}+\sqrt{y+\frac{(z-x)^{2}}{12}}+\sqrt{z+\frac{(x-y)^{2}}{12}} \leq \sqrt{3}
$$
My work -
First, I tried to simplify this and then now I need to prove that $\sqrt{12x+(y-z)^2}+\sqrt{12y+(z-x)^2}+\sqrt{12z+(x-y)^2} < 6 $
which I can't prove.
Then i tried jensen for $f(x)=\sqrt{x}$ and taking weights as $x,y,z$ but it also does not work.
any hints?
thankyou

Since $\frac{1}{12}<\frac{1}{8},$ it's enough to prove that:
$$\sum_{cyc}\sqrt{x+\frac{(y-z)^2}{8}}\leq\sqrt3.$$
Now, by C-S
$$\left(\sum_{cyc}\sqrt{x+\frac{(y-z)^2}{8}}\right)^2\leq\sum_{cyc}\frac{x+\frac{(y-z)^2}{8}}{2x+y+z}\sum_{cyc}(2x+y+z)=\sum_{cyc}\frac{4x+\frac{(y-z)^2}{2}}{2x+y+z}$$ and it's enough to prove that
$$\sum_{cyc}\frac{4x+\frac{(y-z)^2}{2}}{2x+y+z}\leq3$$ or
$$\sum_{cyc}\frac{8x(x+y+z)+(y-z)^2}{2x+y+z}\leq6(x+y+z)$$ or
$$\sum_{cyc}\left(6x-\frac{8x(x+y+z)+(y-z)^2}{2x+y+z}\right)\geq0$$ or
$$\sum_{cyc}\frac{4x^2-y^2-z^2-2xy-2xz+2yz}{2x+y+z}\geq0$$ or
$$\sum_{cyc}\frac{(x-y)(2x+y-z)-(z-x)(2x+z-y)}{2x+y+z}\geq0$$ or
$$\sum_{cyc}(x-y)\left(\frac{2x+y-z}{2x+y+z}-\frac{2y+x-z}{2y+x+z}\right)\geq0$$ or
$$\sum_{cyc}(x-y)^2z(x+y)(2z+x+y)\geq0$$ and we are done!
