Is $x^2 \geq \alpha(\alpha-1)$? 
If $\alpha$ is a nonnegative real and $x$ is a real satisfying $(x+1)^2\geq \alpha(\alpha+1),$ is $x^2 \geq \alpha(\alpha-1)$?

The answer is yes. Consider two cases: $1) \, x < -1$ and $2)\, x\geq -1.$ 
In case $1,$ taking the square root of both sides of the inequality gives $-(x+1) \geq \sqrt{\alpha(\alpha+1)}\Rightarrow x \leq -\sqrt{\alpha(\alpha+1)}-1.$ Hence $x^2\geq \alpha(\alpha+1)+2\sqrt{\alpha(\alpha+1)}+1.$ Since $2\sqrt{\alpha(\alpha+1)}+1\geq 2\alpha+1 > -2\alpha, x^2>\alpha(\alpha+1)-2\alpha=\alpha(\alpha-1).$ 
Now in case $2,$ if $\alpha = 0, x^2 \geq 0,$ so we are done. Suppose $\alpha > 0$. Taking the square root of both sides gives $x+1 \geq \sqrt{\alpha(\alpha+1)}\Rightarrow x\geq \sqrt{\alpha(\alpha+1)}-1\Rightarrow x^2\geq \alpha(\alpha+1)-2\sqrt{\alpha(\alpha+1)}+1.$ $2\alpha-2\sqrt{\alpha(\alpha+1)}+1 =2\alpha\left(1-\sqrt{1+\frac{1}{\alpha}}\right)+1 =\dfrac{\sqrt{1+\frac{1}{\alpha}}-1}{1+\sqrt{1+\frac{1}{\alpha}}}>0.$ Hence $-2\sqrt{\alpha(\alpha+1)}+1>-2\alpha\Rightarrow x^2 > \alpha(\alpha+1)-2\alpha = \alpha(\alpha-1)$. I think this argument works, but I was wondering if there was a faster method? 

Here is one way by contraposition that has some advantage in not requiring cases. We can assume $\alpha > 1$ otherwise $\alpha(\alpha-1) \leq 0 \leq x^2$. (This is so we can take the square root.)
Suppose $x^2 < \alpha(\alpha-1)$, so 
$$ (x + 1)^2 = x^2 + 2x + 1 < \alpha(\alpha - 1) + 2\sqrt{\alpha(\alpha-1)} + 1. $$
It suffices to show this is less than $\alpha(\alpha + 1)$. Expanding, this amounts to showing 
$$ -\alpha + 2\sqrt{\alpha(\alpha-1)} + 1 < \alpha. $$
This is easy, because you can rearrange to get $2\sqrt{\alpha(\alpha-1)} < 2\alpha - 1$ and then square (as both LHS and RHS are positive) to get $4\alpha(\alpha-1) < (2\alpha-1)^2 = 4\alpha(\alpha - 1) + 1$. That is, $x^2 < \alpha(\alpha-1)$ implies $(x + 1)^2 < \alpha(\alpha + 1)$. 
EDIT: I forgot to add that when we write $2x < 2\sqrt{\alpha(\alpha-1)}$ in the first step, this is because $2x \leq 2|x| < 2\sqrt{\alpha(\alpha-1)}$, i.e. we can assume $x$ is nonnegative. 
