Troubles calculating a set infimum

Question

I have a set like this one:

$$A=\left\{\sqrt{x^2+x}-x, x \in \Re\right\}$$

I am trying to calculate the infimum, which seeing a plot drawn with wolfram has been revealed to be zero (for x=0).

So I try to calculate if this inequality may be true for all values of x in some interval:

$$\sqrt{x^2+x}-x \ge d $$

It becomes:

$$x(1-2d) - d^2 \ge 0 $$ $$\Delta(d)= 4d^2 -4d +1$$

So it has a lower bound if ∆(d)<0 for some d values, for all x values. But ∆(d) may never be negative, it has only a zero. So according to this solution the set doesn't have a lower bound, but it has one !! What did I do wrong here?

PS: I am following a book which solves these problems without using a precise rule. What I need is the rule: is there a rule which says how to compute if a function has a lower/upper bound, and in case the infimum/supremum? I just see tricky ways to solve these problems, which I often understand, but I don't understand what pattern is followed to solve them, I don't see a precise algorithm. If someone knows material/books which explain this please post them, I haven't found anything of useful on the web.

PS: I need to solve it without using limits and derivatives, just with the method I've shown, if possible.

Answer 1

The general rule to solve these kind of exercises is through the study of the function (limits, derivative, etc). However generally these exercises are carried out before one sees limits and derivatives...

In this particular case one understand that the function goes to zero as x->0 while it is clearly always positive.

Answer 2

First note this, the function is defined on the domain $ (-\infty,-1] \cup [0,\infty)$. Also you can see this $f(0)=0$, which what you want to prove, and $ f(-1)=1. $ Now, prove that the function is increasing on the interval $[0,\infty)$ and decreasing on $(-\infty,-1]$. Also, note that

$$ \lim_{x\to \infty}\sqrt{x^2+x}-x=\frac{1}{2},\, \lim_{x\to -\infty}\sqrt{x^2+x}-x = \infty. $$