Given two natural numbers $p$ and $i$, such that $0 < i \leqslant 2^p$, let
$$
\Phi(p,i) := \frac{1}{2^p+1}
+ \frac{1}{(i+1)^2} - \frac{1}{2^p}\lg\left(\frac{2^p}{i}+1\right),
$$
where $\lg x$ is the binary logarithm. With the help of a Computer Algebra System, it *seems* that


* If $0 \leqslant p \leqslant 3$, then $\Phi(p,i) < 0$.
* If $4 \leqslant p$, there exists $i\_p$ such that $\Phi(p,i\_p) = 0$
and $\Phi(p,i) > 0$ for $1 \leqslant i < i\_p$,
and $\Phi(p,i) < 0$ for $i\_p < i \leqslant 2^p$.



> 
> How can I prove this?
> 
> 
> 


Just in case, the partial derivative with respect to $i$ is:
$$
\frac{\partial\Phi}{\partial i}(p,i) = \frac{1}{i(2^p+i)\ln 2} - \frac{2}{(i+1)^3},
$$
where $\ln x$ is the natural logarithm.


[Note: I asked this question over at <https://math.stackexchange.com/> but received no answer nor comments.]



Root and sign of a complicated bivariate function
Suppose $p\ge3$, so the cubic $q(i)$ has $3$ real roots. As the product of the roots is $-1$, at most two of them are positive. From $q(0)=1>0$, $q(1)<0$, $q(+\infty)>0$ we see that there are exactly two positive roots, one in $]0,1[$, the other one in $]1,\infty[$. Now $\Phi(0+)=-\infty$, $\Phi(1)>0$, $\Phi(2^p)<0$, and the result follows from the intermediate theorem.


