inequality $\frac{1}{1+3a}+\frac{1}{1+3b}+\frac{1}{1+3c}+\frac{1}{1+3d} \geq 1$ Given real numbers $a, b, c, d$ with this condition: $$abcd=1$$
Let's prove this inequality: $$\frac{1}{1+3a}+\frac{1}{1+3b}+\frac{1}{1+3c}+\frac{1}{1+3d} \geq 1$$
Thank everybody!
 A: The function $f(x ) = \frac{1}{1+3 e^x}$ is convex on $[\ln(\frac{1}{3}),\infty)$ look: convexity proof.
Whenever $y\leq 1/3$ then we have that $\dfrac{1}{1+3y}   \geq \dfrac{1}{2}$, therefore whenever more than two of the $a,b,c,d$ are less than $1/3$ then the inequality holds. So, it remains to deal with the case that all $a,b,c,d$ are more  than $\frac{1}{3}$ or only one of them is less than $1/3$.
We now deal with the first case, and convexity gives 
$$ f(\ln a) + f(\ln b)+ f(\ln c)+f(\ln d) \geq 4 f\left( \frac{1}{4}\ln (abcd) \right) =4f(0)= 1$$
Now, for the second case assume that $d\leq 1/3$ and $a,b,c >1/3$. We use Jensen's again
$$ f(\ln a) + f(\ln b)+ f(\ln c) \geq 3 f\left( \frac{1}{3}\ln (abc) \right) =            3 f\left( \frac{1}{3}\ln (1/d) \right) = \frac{3}{1+3d^{-\frac{1}{3}}}$$
Hence, we need to show 
$$  \frac{1}{1+3d}+  \frac{3}{1+3d^{-\frac{1}{3}}}\geq 1  $$
This case is dealt in M. Roseberg's post. 
A: Given inequality is wrong. Try $a\rightarrow-\frac{1}{3}^-$.
For positive variables by Vasc's RCF Theorem it's enough to prove starting inequality for 
$b=c=a$ and $d=\frac{1}{a^3}$, which gives $(a-1)^2(a+2)\geq0$.
Done!
