Entropy rates, inverse branches and shifted root contracts

For $a>0$, put $I(a)=a\log a-a+1$. Then $I\mathrm{\prime }(a)=\log a$, $I\mathrm{\prime }\mathrm{\prime }(a)=1\mathrm{/}a>0$, $I(1)=0$, $I(a)\to 1$ as $a\downarrow 0$, and $I(a)\to \mathrm{\infty }$ as $a\to \mathrm{\infty }$. Its lower inverse branch $a_{-}:(0,1)\to (0,1)$ and upper inverse branch $a_{+}:(0,\mathrm{\infty })\to (1,\mathrm{\infty })$ are smooth with $a_{\pm }\mathrm{\prime }(t)=1\mathrm{/}\log a_{\pm }(t)$. In $I(1+c+s)=t$, require $1+c+s>0$ and choose a branch, giving $c=a_{\pm }(t)-1-s$. Under $0<\epsilon <1$ and $1+\epsilon <\gamma <1+{\epsilon }^{-2}$, define $c_{\pm }(\gamma )=a_{\pm }(1\mathrm{/}\gamma )-1$ and $c_{\pm }(\gamma ,\epsilon )=a_{\pm }((1+\epsilon )\mathrm{/}\gamma )-1\mp \epsilon$. Then $-1<c_{-}(\gamma )<c_{-}(\gamma ,\epsilon )<0<c_{+}(\gamma ,\epsilon )<c_{+}(\gamma )$. For fixed $\gamma >1$ the shifted roots tend to the unshifted roots as $\epsilon \downarrow 0$. In particular $1+{\epsilon }^{-2}<(1+\epsilon )\mathrm{/}I(1-\epsilon )$. Different parameters or branches cannot be ordered without the corresponding right-hand sides and shifts.