How to find the minimum value of this function? How to find the minimum value of 
$$\frac{x}{3y^2+3z^2+3yz+1}+\frac{y}{3x^2+3z^2+3xz+1}+\frac{z}{3x^2+3y^2+3xy+1}$$,where $x,y,z\geq 0$ and $x+y+z=1$.
It seems to be hard if we use calculus methods. Are there another method? I have no idea.
Thank you.
 A: This is again a question of the symmetric type, such as listed in:

Why does Group Theory not come in here?

With a constraint $\;x+y+z=1\;$ and $\;x,y,z > 0$ . Sort of a general method to transform such a constraint into the inside of a triangle in 2-D has been explained at length in:

How prove this inequality $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$

The above preliminaries are completed with:

Do symmetric problems have symmetric solutions?
From the latter article comes the following
Theorem (The Purkiss Principle). Let $f$ and $g$ be symmetric functions with continuous second derivatives in the neighborhood of a point $P = (r, \cdots, r)$. On the set where $g$ equals $g(P)$, the function $f$ will have a local maximum or minimum at $P$ except in degenerate cases.

Picture on the left: geometry of the conditions $\;x,y,z > 0\;$ and $\;x+y+z=1$ .
Picture on the right: contour lines of $f(x,y,z)$ , as seen in the plane of the $\color{red}{red}$ triangle, are at $20$ equidistant levels, between the minimum and the maximum as found within the viewport
Our function $f$ in this case is:
$$ f(x,y,z) = \frac{x}{3y^2+3z^2+3yz+1}+\frac{y}{3x^2+3z^2+3xz+1}+\frac{z}{3x^2+3y^2+3xy+1} $$
Applying the Purkiss Principle gives:
$$ g(r,r,r)=r+r+r=1 \quad \Longrightarrow \quad r=\frac{1}{3} \quad \Longrightarrow \quad f(r,r,r) = \frac{1}{2} $$
The main problem with the Purkiss Principle, most of the time, is to prove that the extreme found is global.
Or at least global enough , i.e. an absolute extreme inside our ($\color{red}{red}$) triangle.
In our case, the function will become less than $1/2$, namely close to zero, for $\;|x|,|y|,|z|\rightarrow \infty$ .
So it's clear that the Purkiss Principle is "violated", outside the triangle at least.
These regions, where $\;f(x,y,z)\;$ is less than, say, $1/2 + 0.001$ , are colored $\color{blue}{blue}$ in the picture on the right.
The $\color{blue}{blue\, spot}$ in the picture on the right is a proof without words that the only minimum inside the triangle is (indeed $= 1/2$ ) and at its center $(x,y,z) = (1/3,1/3,1/3)$ . This finishing touch is thus an informal proof (: disclaimer).

Update : a few further details.

Function values close to viewport minimum are white
Function values close to viewport maximum are black
Contour grey values are just the other way around
Function values at the corners of the triangle $=1$ (black)
Function values at the edge centers of the triangle $=4/7$

A: To find a minimum of the function I'm going to find a lower bound and show that the lower bound is attained. For that I'm going to use nothing harder than Cauchy-Schwarz inequality.
Let $f(x, y, z)$ denote your function. Using mentioned Cauchy-Schwarz inequality we get:
$$f(x, y, z)\cdot\big(x(3y^2+3z^2+3yz+1) + y(3x^2+3z^2+3xz+1) + z(3x^2+3y^2+3xy+1)\big) \\
\geqslant (x+y+z)^2 = 1.$$
So,
$$f(x, y, z) \geqslant \frac{1}{3(xy^2 + xz^2 + yx^2 + yz^2 + zx^2 + zy^2 + 3xyz) + x+y+z} \\
=\frac{1}{3(x+y+z)(xy+yz+xz) + 1} = \frac{1}{3(xy+yz+xz) + 1}$$
We want to get an upper bound of $xy + yz + xz$ under the condition $x + y + z = 1$. We can do this, for example:
$$\begin{align}
xy + yz + xz &\leqslant x^2 + y^2 + z^2 \\
3(xy + yz + xz) &\leqslant (x + y + z)^2 = 1 \\
xy + yz + xz &\leqslant \frac{1}{3}
\end{align}$$
Finally,
$f(x, y, z) \geqslant \dfrac{1}{3\cdot\frac{1}{3} + 1} = \dfrac{1}{2}.$ So $\dfrac{1}{2}$ is the lower bound.
Since $\displaystyle f\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2}$, we can see that we've found a point where the lower bound is attained so this is the minimum of the function.
A: By C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{x}{3y^2+3z^2+3yz+1}=\sum_{cyc}\frac{x}{3y^2+3z^2+3yz+(x+y+z)^2}=$$
$$=\sum_{cyc}\frac{x}{x^2+4y^2+4z^2+2xy+2xz+5yz}=$$
$$=\sum_{cyc}\frac{x^2}{x^3+4xy^2+4xz^2+2x^2y+2x^2z+5xyz}\geq$$
$$\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(x^3+6x^2y+6x^2z+5xyz)}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(2x^3+6x^2y+6x^2z+4xyz)}=$$
$$=\frac{(x+y+z)^2}{2(x+y+z)^3}=\frac{1}{2}.$$
The equality occurs for $x=y=z=\frac{1}{3}$, which says that the answer is $\frac{1}{2}$.
Done!
A: $\textbf{Hint:}$ This is not hard using calculus methods, you can use Lagrange Multipliers.
In a nutshell it says that, to find the maximum and minimum value for $f(x,y,z)$ subject to $g(x,y,z)=k$ you find all $x,y,z$ and $\lambda$ such that:
$$\begin{cases} \nabla f(x,y,z) = \lambda \nabla g(x,y,z) \\g(x,y,z) = k \end{cases}$$
Then you evaluate $f$ in such points, the maximum value you get is the maximum value for the function, the minimum value you get is the minimum value for the function.
Remember that $\nabla f(x,y,z) = (f_x,f_y,f_z)$.
Almost always the "tricky" part in problems involving Lagrange Multipliers is to solve the above system, which seems not to be the case in this example.
A: I used wolfram alpha: http://goo.gl/Cjzigc

Looks like the min is at $(1/3,1/3,1/3)$.
