inequality $\frac{1}{1+3a}+\frac{1}{1+3b}+\frac{1}{1+3c}+\frac{1}{1+3d} \geq 1$

Question

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!

Answer 1

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.