<s>[INST] Proving that $\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 } $ 
Let $x,y,z\geq 1$ and $x+y+z=6$. Then $$\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 }. $$

I tried to use Cauchy- Schwartz inequality but it doesn't work.
 [/INST] By Jensen's inequality with  $\phi\left(t\right)=\frac{1}{3t^2+1}$ we have 
$$\phi\left(\frac13\sum_{i=1}^{3}x_i\right)\le \frac13\sum_{i=1}^{3}\phi\left(x_i\right) $$
since $x+y+z= 6$ we get 
$$\frac{1}{13}=\phi\left(2\right)\le \frac13\left(\frac{1}{3x^2+1}+\frac{1}{3y^2+1}+\frac{1}{3z^2+1}\right) $$
hence $$\left(\frac{1}{3x^2+1}+\frac{1}{3y^2+1}+\frac{1}{3z^2+1}\right)\ge \frac{3}{13}\ge \frac{3}{16} $$
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