= Solution
Take $r>0$ and $m>0$, the positive-energy sigma-model regime, and keep $\theta=0$ for the <energy> calculation. For a static field the <energy> is
$$
E=\frac r4\int_{\mathbb R}\left[\phi_x^2+\sin^2\phi\,\alpha_x^2+m^2\sin^2\phi\right]dx.
$$
For either sign $s=\pm1$, complete the square:
$$
\begin{aligned}
E&=\frac r4\int\left[(\phi_x-sm\sin\phi)^2+\sin^2\phi\,\alpha_x^2\right]dx
+\frac{rm s}{2}\int\phi_x\sin\phi\,dx\\
&=\frac r4\int\left[(\phi_x-sm\sin\phi)^2+\sin^2\phi\,\alpha_x^2\right]dx+rm sT.
\end{aligned}
$$
Here $\int\phi_x\sin\phi\,dx=-[\cos\phi]_{-\infty}^{+\infty}=2T$. Choose $s$ to have the sign of $T$. The <Bogomolny bound> is therefore
$$
\boxed{E\geq rm|T|.}
$$
Saturation requires $\phi_x=sm\sin\phi$ and $\sin\phi\,\alpha_x=0$. Finite-energy vacua have $\sin\phi=0$, so their endpoint cosine values are $\pm1$ and this particular <topological charge> is $T\in\{-1,0,1\}$. This is the <static kink bound in a mass-deformed spherical sigma model>. Negative $r$ would instead make the derivative <energy> unbounded below, so positivity is an essential physical convention.
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