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.

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.
