<s>[INST] $ \frac{1}{3a^2+1}+\frac{1}{3b^2+1}+\frac{1}{3c^2+1}+\frac{1}{3d^2+1} \geq \frac{16}{7}$ Let $a,b,c,d >0$ and $a+b+c+d=2$. Prove this:
$$ \frac{1}{3a^2+1}+\frac{1}{3b^2+1}+\frac{1}{3c^2+1}+\frac{1}{3d^2+1} \geq \frac{16}{7}$$
 [/INST] Let $a=\frac{x}{2}$, $b=\frac{y}{2}$, $c=\frac{z}{2}$ and $d=\frac{t}{2}$.
Hence, $x+y+z+t=4$ and we need to prove that $\sum\limits_{cyc}\frac{1}{3x^2+4}\geq\frac{4}{7}$.
Let $f(x)=\frac{1}{3x^2+4}$.
Hence, $f''(x)=\frac{6(9x^2-4)}{(3x^2+4)^2}>0$ for all $1\leq x\leq4$.
Thus, by Vasc's RCF Theorem it's enough to prove our inequality 
for $z=y=x$ and $t=4-3x$, where $1\leq x\leq\frac{4}{3}$.
Id est, we need to prove that
$$\frac{3}{3x^2+4}+\frac{1}{3(4-3x)^2+4}\geq\frac{4}{7},$$
which is $(x-1)^2(3x+2)(4-3x)\geq0$.
Done!
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