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.
 A: 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 $$\left(\frac{1}{3x^2+1}+\frac{1}{3y^2+1}+\frac{1}{3z^2+1}\right)\ge \frac{3}{13}\ge \frac{3}{16} $$
A: Using Lagrange multipliers, write
$$F(x,y,z) = \frac{1}{3x^2+1}+\frac{1}{3y^2+1}+\frac{1}{3y^2+1}-\lambda(x+y+z)$$  Taking derivatives, we get $$\frac{6x}{(3x^2+1)^2}=\frac{6y}{(3y^2+1)^2}=\frac{6z}{(3z^2+1)^2}=-\lambda$$
Now it's straightforward to show that $f(t) = \frac{6t}{(3t^2+1)^2}$ is one-to-one on $t>1,$ so $x=y=z=2.$ 
A: Also, the Tangent Line method works: $$\sum_{cyc}\frac{1}{3x^2+1}=\sum_{cyc}\left(\frac{1}{3x^2+1}-\frac{1}{13}+\frac{12}{169}(x-2)\right)+\frac{3}{13}=$$
$$=\sum_{cyc}\frac{3(x-2)^2(12x+11)}{169(3x^2+1)}+\frac{3}{13}\geq\frac{3}{13}>\frac{3}{16}.$$
A: 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 $$\frac1{3\cdot2^2+1}+\frac1{3\cdot 2^2+1}+\frac1{3\cdot 2^2+1}=\frac3{13}\ge \frac3{16}$$.
