<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] 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 $x=y=z=2.$ 
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