<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] The problem is symmetric in $x,y$ and $z$ ;  and the "feasible" region is convex.. .  $\therefore $ the minimum will occur when $x=y=z$, or else at a boundary point... 
but on the boundary at least one of $x,y,z$ is $1$, and we easily get a lower bound of $\frac14$, coming from that term...
Finally if $x=y=z=2$
we get $$\frac1{3\cdot2^2+1}+\frac1{3\cdot 2^2+1}+\frac1{3\cdot 2^2+1}=\frac3{13}\ge \frac3{16}$$.
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