Find a lower bound of $-\frac{1}{4}\sqrt{(1-2a+x)^2-8(-3-4a-3x-2ax)}-\frac{1}{4}(1-2a+x)$ I have to find a lower bound (that does not depend on $x$) of the following quantity:
$$-\frac{1}{4}\sqrt{(1-2a+x)^2-8(-3-4a-3x-2ax)}-\frac{1}{4}(1-2a+x)$$
where $x\geq 0,\, a>0$.
Really I have tried to observe that:
$-\frac{1}{4}\sqrt{(1-2a+x)^2-8(-3-4a-3x-2ax)}-\frac{1}{4}(1-2a+x)>-\frac{1}{2}\sqrt{(1-2a+x)^2-8(-3-4a-3x-2ax)}=A$
but then I can't find a lower bound of A...can you help me?

Let $a$ be a fixed positive real. Let $x \geqslant 0$ be allowed to vary.
Let $A=2,\quad B=1-2a+x,\quad C=-3-4a-3x-2ax.$
Consider the quadratic $Ay^2+By+C=0.$
Then your expression is the left-hand root of this quadratic:
$$-\frac{1}{4}\sqrt{(1-2a+x)^2-4A(-3-4a-3x-2ax)}-\frac{1}{4}(1-2a+x).$$
(For we see that $A>0,\; C<-3,\; B^2-4AC>0$ for any $x$, and thus the parabola always has two roots.)
So I think you cannot have such a bound. For we can, by increasing $x$ more and more, move the axis of symmetry of the parabola further and further to the left. This must in turn push the left-hand root towards $-\infty$.
The axis of symmetry:  $\;y= \frac{-B}{2A}.$
