I am studying a research paper in analytic number theory
It involves an inequality which is to be used in sharpening a result. Unfortunately I am unable to see how to get the required inequality (See proposition $1$) Or read below
$a$ is odd integer $\geq$3 and $r$ is an integer lying between $1\le r<a/2$ . Author deduced $(12)$
French to English in 2nd line after $(12)$ is
and $s_0$ is the only root in $(0, 1)$ of polynomial $Q(s) =~\dots$
In particular $\delta(a) \geq~\dots$
Now, inequality to be deduced is $(13)$ of image using $$\phi_{r, a}\geq \frac{2^{r+1}} { r^{a-2r}}\text{ and }2r\leq{2r+1}\leq{2(r+1) }.$$
Taking $\log$ both sides in $\phi_{r, a} $ and putting it in $(12)$ I got RHS of $(12)$ equals $$\frac{ (2a-3r-1) \log2 + (2r+1) \log(2r+1) +(a-2r) \log r } { a + (a-2r) \log2+(2r+1)\log(2r+1) } $$ which is not equal to $(13)$.
Can someone please tell how to derive it to $(13)$?
Consider the numerator and denominator of RHS in (12). We have \begin{align} &(a-2r)\log(2) + (2r+1)\log(2r+1) - \log(\varphi_{r,a})\\ \ge\ & (a-r)\log(2) - r\log(2) + (2r+1)\log (2r) - \log(\varphi_{r,a})\\ =\ & (a-r)\log(2) + (r+1)\log(2) + (2r+1)\log(r) - \log(\varphi_{r,a}) \end{align} and \begin{align} &a + (a-2r)\log(2) + (2r+1)\log(2r+1)\\ \le\ & a + 1 + (a-2r)\log(2) + (2r+1)\log (2r+2)\\ =\ & (a+1) + (a+1)\log(2) + (2r+1)\log(r+1). \end{align}
Thus, to prove (13), it suffices to prove that $$(r+1)\log(2) + (2r+1)\log(r) - \log(\varphi_{r,a}) \ge (a+1)\log(r)$$ or $$\frac{2^{r+1}}{r^{a-2r}} \ge \varphi_{r,a}.$$
Consider $Q(s) = rs^{a+2} - (r+1)s^{a+1} + (r+1)s - r$. Note that $Q(0) < 0$, $Q(1)=0$, and \begin{align} Q(1 - \tfrac{1}{r}) = -2 (1-\tfrac{1}{r})^{a+1} - \tfrac{1}{r} < 0. \end{align} Thus, $s_0 > 1 - \frac{1}{r}$.
Now we are ready to prove that $\frac{2^{r+1}}{r^{a-2r}} \ge \varphi_{r,a}$. We have to prove that $$2^{r+1} \ge ((r+1)s_0 - r)^r(r+1-rs_0)^{r+1} (r - rs_0)^{a-2r}. \tag{1}$$ Since $(r+1)s_0 - r < r+1 - r = 1$ and $r+1 - rs_0 \le r + 1 - r(1-\tfrac{1}{r}) = 2$ and $r - rs_0 \le r - r(1 - \tfrac{1}{r}) = 1$, (1) is true.
We are done.