If $\{a,b,c,d,e\}\subset[0,1]$ so $\sum\limits_{cyc}\frac{1}{1+a+b}\leq\frac{5}{1+2\sqrt[5]{abcde}}$ Let $\{a,b,c,d,e\}\subset[0,1]$. Prove that:
$$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+d}+\frac{1}{1+d+e}+\frac{1}{1+e+a}\leq\frac{5}{1+2\sqrt[5]{abcde}}$$
I tried C-S, convexity and more, but without success.
 A: Here are two cases.
Case 1: (cyclically) ${a b} < \frac14$ . 
Observe $a+b \ge 2 \sqrt{a b}$ by AM-GM, likewise for the other terms. So it is enough to prove
$$
\sum_{cyc} \frac{1}{1 +2 \sqrt{a b}} \leq\frac{5}{1+2\sqrt[5]{abcde}}
$$
Let $2 \sqrt{a b} = x$, $2 \sqrt{b c} = y$, etc.
Then we need
$$
\frac 15 \, \sum_{cyc} \frac{1}{1 +x} \leq\frac{1}{1+ \sqrt[5]{\prod_{cyc}x}}
$$
Consider the function 
$$
f(z) = \frac{1}{1 +e^z}
$$
We have 
$$
f''(z) = \frac{e^z (e^z -1)}{(1+e^z)^3}
$$
so for $z < 0$ we have that $f''(z) < 0$ and hence $f(z)$ is strictly concave. Hence by Jensen,
$$
\frac 15 \, \sum_{cyc} f(z_i) \leq f(\frac 15 \, \sum_{cyc} z_i)
$$
Now apply $e^{z_i} = x$ cyclically. This establishes the inequality for $z < 0$, i.e. when  all $x = 2 \sqrt{ab} < 1$. 
Case 2: (cyclically) $a + b \leq 1$ . 
I am grateful for a comment by  Hugh Denoncourt  which lead to this case.
Define a vector $(w,z)$. Consider the function 
$$
f(w,z) = - \frac{1}{1 +e^z + e^w}
$$
which is the negative of the function under consideration, so we are looking for convexity.
We have the second partial derivative:
$$
\frac{\partial^2 f(w,z)}{\partial w^2} = \frac{e^w (1 +e^z - e^w)}{(1+e^z+e^w)^3}
$$
and for $z$ likewise.
We also have the Hessian of this function:
$$
H = \frac{e^z e^w (1 -e^z - e^w)}{(1+e^z+e^w)^5}
$$
so for $e^z + e^w < 1$ we have that $H > 0$ and hence $f(w,z)$ is strictly convex. Hence by Jensen's inequality for vector-valued functions,
$$
\frac 15 \, \sum_{cyc} f(w_i,z_i) \geq f(\frac 15 \, \sum_{cyc} (w_i,z_i))
$$
Now apply $e^{z_i} = a$ and $e^{w_i} = b$ cyclically. This establishes the inequality for all $a+b<1$. However, this will not solve the case fully for $0 \leq a,b \leq 1$. 
Comment: If it were not the Hessian, but positivity of the second partial derivatives, this would always be given for $0 \leq a,b \leq 1$. Do we need the Hessian?
