Find a lower bound of $-\frac{1}{4}\sqrt{(1-2a+x)^2-8(-3-4a-3x-2ax)}-\frac{1}{4}(1-2a+x)$

Question

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?

Answer 1

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: $\;y= \frac{-B}{2A}.$