is $12\frac{6}{7}$ the correct answer?

Yes, the minimum is $\frac{90}{7}$.

I wasn't quite sure about this, but if it's correct, I'll post my proof: From QM-AM: $\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}\geq\frac{x+y+z}{3}\geq2 \Leftrightarrow x^{2}+y^{2}+z^{2}\geq12.$ (This can be also proved by Cauchy-Schwarz's) So, inequality $x^{2}+y^{2}+z^{2}+x+y+z+3\geq21$ is true. (*) Let $B=\frac{x}{y^{2}+z+1}+\frac{y}{z^{2}+x+1}+\frac{z}{x^{2}+y+1}$ Using Chebyshev's we get: $3B\geq(x+y+z)[\frac{1}{y^{2}+z+1}+\frac{1}{z^{2}+x+1}+\frac{1}{x^{2}+y+1}]\geq6\cdot(\frac{1}{y^{2}+z+1}+\frac{1}{z^{2}+x+1}+\frac{1}{x^{2}+y+1})$ Multiplying this with 21 and using (*), we get: $63B\geq6\cdot [(y^{2}+z+1)+(z^{2}+x+1)+(x^{2}+y+1)](\frac{1}{y^{2}+z+1}+\frac{1}{z^{2}+x+1}+\frac{1}{x^{2}+y+1})\geq6\cdot9.$ Last inequality follows from AM-HM applied to the numbers $y^{2}+z+1, z^{2}+x+1$ and $x^{2}+y+1$. So $63B\geq 54 \Leftrightarrow B\geq \frac{6}{7}$. So the required minimum is $12+\frac{6}{7}=\frac{90}{7}$ . It's attained for $x=y=z=2$

ivanbart-15 wrote: I wasn't quite sure about this. Let $B=\frac{x}{y^{2}+z+1}+\frac{y}{z^{2}+x+1}+\frac{z}{x^{2}+y+1}$ $B\geq \frac{6}{7}$. Well, you can't divide at this way, because there is a big mistake, take $x=y=z \to \infty$ and we get $B \to 0$ so its minimum isn't correct. We have to let the $\sum x^2$ by it. The mistake is you use $21\ge \sum x^2+x+1$ , while it is in other way. The first thing that easy is to remark is $x+y+z=6$ when the minimum is, because $x-> x-\delta$ gives $g\le f-\delta$

Yeah, I wasn't sure because of that, although my result is correct. Is my proof adequate, if we take $x+y+z=6$ and $x^{2}+y^{2}+z^{2}+x+y+z=21$?. Does anybody have another proof?

ivanbart-15 wrote: Is my proof adequate, if we take $x+y+z=6$ and $x^{2}+y^{2}+z^{2}+x+y+z=21$?. Yes, we can take $x+y+z=6$, which gives very ugly and smooth proof, but $x^{2}+y^{2}+z^{2}+x+y+z=21$ is not adequate.

See: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2396731#p2396731

arqady wrote: ivanbart-15 wrote: Is my proof adequate, if we take $x+y+z=6$ and $x^{2}+y^{2}+z^{2}+x+y+z=21$?. Yes, we can take $x+y+z=6$, which gives very ugly and smooth proof, but $x^{2}+y^{2}+z^{2}+x+y+z=21$ is not adequate. please you post your solution?

Shu wrote: Let $x$, $y$, $z$ be positive real numbers satisfying $x+y+z\ge 6$. Find, with proof, the minimum value of \[ x^2+y^2+z^2+\frac{x}{y^2+z+1}+\frac{y}{z^2+x+1}+\frac{z}{x^2+y+1}. \] Hint: $\sum \left[ \frac{2(x^2+y+1)}{49}+\frac{z}{x^2+y+1}+\frac{z^2}{14} \right] \ge \frac{3}{7}(x+y+z)$ and $3(x^2+y^2+z^2) \ge (x+y+z)^2$

ivanbart-15 wrote:

Click to reveal hidden text

$x^{2}+y^{2}+z^{2}+x+y+z+3\geq21$ Click to reveal hidden text

$63B\geq6\cdot [(y^{2}+z+1)+(z^{2}+x+1)+(x^{2}+y+1)](\frac{1}{y^{2}+z+1}+\frac{1}{z^{2}+x+1}+\frac{1}{x^{2}+y+1})\geq6\cdot9.$ Click to reveal hidden text

$x^{2}+y^{2}+z^{2}+x+y+z+3\geq21$ $21\geq (y^{2}+z+1)+(z^{2}+x+1)+(x^{2}+y+1).$ Mistake.

arqady wrote: Yes, we can take $x+y+z=6$, which gives very ugly and smooth proof, but $x^{2}+y^{2}+z^{2}+x+y+z=21$ is not adequate. why we can take $x+y+z=6$ ? can you explain more ?