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? 
 A: Let $\Delta=(x+1)^2-\alpha(\alpha+1)$, so that $\Delta\geq 0$. Observe that
$$
x^2-\alpha(\alpha-1)=\Delta+2(\alpha-x)-1,
$$
which is non-negative whenever $x\leq \alpha-\tfrac12$.
On the other hand
$$
x^2-\alpha(\alpha-1)=x^2-(\alpha-\tfrac12)^2+\tfrac14,\qquad (\star)
$$
thus in the remaining case that $x> \alpha-\tfrac12$ one has that either $\alpha>\tfrac12$ and thus we are allowed to square both sides, yielding $x^2> (\alpha-\tfrac12)^2$, or $\alpha\leq \tfrac12$ and thus (by non-negativity of $\alpha$) $\tfrac14\geq (\alpha-\tfrac12)^2$. In either case $(\star)$ is non-negative.
A: Here's another method.
Suppose $\alpha(\alpha+1) = (x+1)^2$ and $x\geq 0$. Then since $\left(x+\frac{1}{2}\right)\left(x+\frac{3}{2}\right)=(x+1)^2-\frac{1}{4},$ we have that $\alpha > x+\frac{1}{2}.$ Hence $\alpha(\alpha-1)=\alpha(\alpha+1)-2\alpha <(x+1)^2-2x-1=x^2.$
If $\alpha(\alpha+1) < (x+1)^2$ and $x\geq 0,$ then take $\beta > \alpha$ with $\beta(\beta + 1) = (x+1)^2.$ Since $\alpha - 1<\beta - 1$ and $\alpha \geq 0,$ $\alpha(\alpha-1) \leq \alpha(\beta-1) <\beta(\beta -1) x^2$.
When $x<0, -|x| -1 < x+1 < |x| + 1,$ so $(x+1)^2 < (|x|+1)^2,$ and $x^2 = |x|^2,$ so the result follows follows from the result for $\alpha, |x|$.
A: 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. 
