Does there exist any strictly decreasing function $f: \Bbb R \to [0,+\infty)$ such that $f(x)-\sqrt{f(x)}$ is strictly increasing function on the whole $\Bbb R$?
Now, we had calculus at high school, but, at the moment, we haven't gone through it in analysis lectures, so, on the first exam we are supposed to use things mentioned in lectures only.
I tried to use the fact that if $f(x)$ is a monotonically decreasing function, then the composition $$f(f(x))$$ is monotonically increasing. I might have chosen the wrong path. I also considered inverses, but it didn't help me either.
Since I have a sum, I assumed, without deriving, I can't look at it as a real composition.
$$\operatorname{Ran}(f)=[0,+\infty)$$ implies $\sqrt{f(x)}$ would be defined well.
I constantly kept returning to some decreasing exponential function $g(x)=a^{x}, a\in(0,1)$ although a rational function with odd exponents seemed equally useful/useless…
I thought that the degree of $f(x)$ must be less than the degree of $\sqrt{f(x)}$ in terms of that exponential function. I haven't proven it.
The role of the leading coefficient also seemed irrelevant in this case, it reminded me of polynomials, but they can't be injective and always positive at the same time globally, without restriction, am I right?
I "concluded" there is no such function that satisfies the constraints.
Best Answer
Proof. Let $a>0$. We have to show that $g(x+a) > g(x)$ for each $x$. This is equivalent to $$ h(x)-h(x+a) > h(x)^2-h(x+a)^2 = (h(x)-h(x+a))(h(x)+h(x+a)). $$ Since $h(x)-h(x+a) > 0$, we may divide and obtain the equivalent inequality $$ h(x)+h(x+a) < 1, $$ which is true by assumption.
In your case, set $h = \sqrt f$. So, whenever $f(x) < \frac 14$ for all $x$, this is a function as desired.