If $\{a,b,c,d,e\}\subset[0,1]$ so $\sum\limits_{cyc}\frac{1}{1+a+b}\leq\frac{5}{1+2\sqrt[5]{abcde}}$

Solution 1:

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?