Let $f: \mathbb R \to \mathbb R$ be a differentiable function such that $$f(x)^2+f'(x)\le 0.$$ Show that the zero function is the only solution.
I started bwoc assuming that there is a point $x$ such that $f(x)$ is not zero. Hence there is an interval such that $f$ is not zero by continuity. Then I rewrite the equation as $(1/f)'(x)\le 1$.
I would prefer if someone could continue my argument and not find another approach. My question is more of "Is this a step in the right direction?" A different approach is welocomed too of course if what i said above is not possible.
First, consider the following problem: We have some function $g\colon (a,b)\to\mathbb{R}$ and we know that $g'(x)\le1$ for $x\in(a,b)$, and $g(x_0)=g_0\neq0$ for some $x_0\in(a,b)$.
Now can we conclude that $g(x_0+t) \le g_0+t$? Yes we can!
Suppose that $g(x_0+t)> g_0+t$, then by the mean value theorem, there is some $c\in(x_0,x_0+t)$ such that $g'(c) = \frac{g(x_0+t)-g_0}{t}$. But then $g'(c)>1$, so we know that $g(x_0+t)\le g_0+t$ for all $t$ in the interval $(a-x_0,b-x_0)$.
Now we can get back to our problem: We obtain that $(1/f)(x_0+t)\le (1/f_0)+t$, and thus $f(x_0+t)\ge\frac1{t+1/f_0}$. But then we have a problem as $t\to-1/f_0$, so the only conclusion is that such a $f$ cannot be defined on all of $\mathbb{R}$.