<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] Also, the Tangent Line method works: $$\sum_{cyc}\frac{1}{3x^2+1}=\sum_{cyc}\left(\frac{1}{3x^2+1}-\frac{1}{13}+\frac{12}{169}(x-2)\right)+\frac{3}{13}=$$
$$=\sum_{cyc}\frac{3(x-2)^2(12x+11)}{169(3x^2+1)}+\frac{3}{13}\geq\frac{3}{13}>\frac{3}{16}.$$
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