I'm reading Morawetz's Two $L^p$ Inequalities, in which two global inequalities for $p>1$ are presented. Here $\lVert \cdot \rVert$ is an $L^p$ norm.
$$\frac{\lVert X\rVert^p + \lVert Y \rVert^p}{2}-\left\lVert \frac{X + Y}{2} \right\rVert^p \ge a \left\lVert \frac{X - Y}{2} \right\rVert^{p/s} \left( \frac{\lVert X\rVert^p + \lVert Y \rVert^p}{2} \right)^{1-1/s}, \tag{Ia}\label{Ia}$$
where $a$ only depends on $p$ and is strictly greater than one,
$$s = \begin{cases} 1 &\text{ for } p < 2 \\ p/2 &\text{ for } p \ge 2, \end{cases} $$
$$\frac{\lVert X\rVert^p + \lVert Y \rVert^p}{2} \le \left\lVert \frac{X + Y}{2} \right\rVert^p \left( 1 + b_1 \left( \frac{\lVert X - Y \rVert}{\lVert X + Y \rVert} \right)^2 + b_2 \left( \frac{\lVert X - Y \rVert}{\lVert X + Y \rVert} \right)^p \right), \tag{Ib}\label{Ib}$$
where $b_1$ and $b_2$ are constants depending only on $p$. Then four variables are introduced
\begin{align} A &= \frac{X + Y}{2} \\ D &= \frac{X - Y}{2} \\ r &= \frac{\lVert D \rVert}{\lVert A \rVert} \\ m &= \left( \frac{\lVert X\rVert^p + \lVert Y \rVert^p}{2} \right)^{1/p} \end{align}
so as to give a range of the ratio $\lVert A \rVert/m$. In the article, it's claimed that
$$(1+b_1r^2+b_2r^p)^{1/p} \le \lVert A \rVert/m \le (1-(cr\lVert A \rVert/m)^{p/s})^{1/p},$$
where $c = c(p) <1$.
However I got $-1/p$ instead of $1/p$ for the power on the LHS and $c = a^{s/p} > 1$ instead of $c < 1$ on the RHS. Have I made any errors?
Equation \eqref{Ia} can be simplified to
\begin{align} m^p - \lVert A \rVert^p &\ge a \lVert D \rVert^{p/s} m^{p(1 - 1/s)} \\ 1 - (\lVert A \rVert/m)^p &\ge a (r \lVert A \rVert/m)^{p/s} \\ 1 - a (r \lVert A \rVert/m)^{p/s} &\ge (\lVert A \rVert/m)^p \\ (1 - a (r \lVert A \rVert/m)^{p/s})^{1/p} &\ge (\lVert A \rVert/m). \end{align}
Equation \eqref{Ib} can be simplified to
\begin{align} m^p &\le \lVert A \rVert^p (1 + b_1r^2 + b_2r^p) \\ (1 + b_1r^2 + b_2r^p)^{-1/p} &\le \lVert A \rVert/m. \end{align}
Thank you in advance for any comments or suggestions.