Knowing that $x+y+z=3$ and $x,y,z\ge0$, how to prove the following inequality? Let $x,y,z\ge 0$, if $x+y+z=3$, show that
$$\dfrac{1}{\sqrt{x^2+xy+y^2}}+\dfrac{1}{\sqrt{y^2+yz+z^2}}+\dfrac{1}{\sqrt{z^2+zx+x^2}}\ge \dfrac{12+2\sqrt{3}}{9}$$
Using this  inequality:
$$\left(\sum_{cyc}\dfrac{1}{\sqrt{x^2+xy+y^2}}\right)\cdot\left(\sum_{cyc}\sqrt{x^2+xy+y^2}\right)\ge 9$$
it is enough to check that
$$\dfrac{9}{\sum_{cyc}\sqrt{x^2+xy+y^2}}\ge \dfrac{12+2\sqrt{3}}{9}$$
And I can't go further.I can't observer something,only find this inequality $=$ iff $x=y,z=0$
 A: Let $g(u,v)=u^{2}+v^{2}+uv$.
Let $f(x,y,z)=g^{-\frac{1}{2}}(x,y)+g^{-\frac{1}{2}}(x,z)+g^{-\frac{1}{2}}(y,z)$ which simply describes the sums you are interested in.
The basic idea here is that we will minimize the function $f(x,y,z)$ constrained to $x+y+z=3$ and show that the absolute minimum is precisely $\frac{12+2\sqrt{3}}{9}$.
So simply compute a partial,
$\quad \quad \quad \quad \quad \quad \frac{\partial f}{\partial x} = -(\frac{x+\frac{y}{2}}{g^{\frac{3}{2}}(x,y)}+\frac{x+\frac{z}{2}}{g^{\frac{3}{2}}(x,z)})$
Now since we are confined to $x,y\geq0$, we know that the leftmost term and rightmost term are positive semi-definite, so the whole expression (because of the minus in front) is negative semi definite. Thus for this partial derivative to be equal to 0, it must be that $x=y=z=0$. But this can't happen because this would imply that $x+y+z=0\neq3$. A local extremum must have all partials equal to zero, so clearly that can't happen if $\frac{\partial f}{\partial x}\neq 0$ (and in fact, by the symmetry of the function, you'd get the same contradiction for the other partials). Therefore, the only critical points within this region must lie on the boundary of the region defined by $x+y+z=3$. These are the three lines defined by
\begin{align*}
z=0, y=3-x, x\in[0,3] 
\end{align*}
\begin{align*}
y=0, z=3-x, x\in[0,3]
\end{align*}
\begin{align*}
x=0, z=3-y, y\in[0,3]
\end{align*}
Again, by the symmetry, I'll just look at one of these lines, specifically $z=0, y=3-x$. Thus we are now looking at one dimensional function $h(x)=f(x, 3-x, 0)$, which through some simplication leads to $h(x)=\frac{1}{x}+\frac{1}{3-x}+\frac{1}{\sqrt{x^{2}-3x+9}}$. From here, we attempt to find extrema for $h(x)$, and I'll skip the computational details of teh derivatives. If you take a derivative, you can see easily that $\frac{3}{2}$ is a solution to make $h'(x)=0$. Then when you compute $h''(x)$, you will see that it is positive definite, meaning $h'(x)$ is strictly increasing, so $x=\frac{3}{2}$ is the ONLY solution to $h'(x)=0$. Now since $h''(x)$ is strictly positive, you now know that the ABSOLUTE minimum on this line must be at $\frac{3}{2}$, and note that $h(\frac{3}{2}$)=$\frac{12+2\sqrt{3}}{9}$. Thus by the previous argument, since it's an absolute min on the boundary, it's an absolute min over the entire surface. You can check for yourself that the other two parts of the boundary give you the exact same result. Thus by definition of absolute minimum, the result you wanted has been proven.
