<s>[INST] 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.
 [/INST] 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!
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