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 http://math.stackexchange.com/ but received no answer nor comments.]