Suppose $\hat{\theta}_u$ is unbiased for $\theta$, and $\hat{\theta}_b = (1+\alpha)\hat{\theta}_u$.
I have already shown that $$\text{MSE}(\hat{\theta}_b) = (1+\alpha)^2\text{MSE}(\hat{\theta}_u)+\alpha^2\theta^2$$ and that $$\text{MSE}(\hat{\theta}_b) < \text{MSE}(\hat{\theta}_u) \Longleftrightarrow \dfrac{-2\text{MSE}(\hat{\theta}_u)}{\text{MSE}(\hat{\theta}_u)+\theta^2} < \alpha < 0\text{.}$$ However, the final solution I am provided is as follows: $$\text{MSE}(\hat{\theta}_b) < \text{MSE}(\hat{\theta}_u) \Longleftrightarrow-2<\dfrac{-2\text{MSE}(\hat{\theta}_u)}{\text{MSE}(\hat{\theta}_u)+\theta^2} < \alpha < 0\text{.}$$ How do we obtain the $-2$ bound for $\alpha$?
My attempt: The only way that $$-2 < \dfrac{-2\text{MSE}(\hat{\theta}_u)}{\text{MSE}(\hat{\theta}_u)+\theta^2}$$ is if $$\dfrac{\text{MSE}(\hat{\theta}_u)}{\text{MSE}(\hat{\theta}_u)+\theta^2} > 1$$ or $$\text{MSE}(\hat{\theta}_u) > \text{MSE}(\hat{\theta}_u)+\theta^2$$ which doesn't seem to be true... because $\theta^2 > 0$, so if anything, the inequality should be reversed. What am I missing here?
Since $\mathrm{MSE}(\hat\theta_u)$ and $\theta^2>0$, you certainly have $\mathrm{MSE}(\hat\theta_u)\big/(\mathrm{MSE}(\hat\theta_u)+\theta^2)\le1$ so $-1\le-\mathrm{MSE}(\hat\theta_u)\big/(\mathrm{MSE}(\hat\theta_u)+\theta^2)$ so $$ -2 \le \frac{-2 \mathrm{MSE}(\hat\theta_u)}{\mathrm{MSE}(\hat\theta_u)+\theta^2}.$$