Motivation behind proof

The following problem is taken from IMC $2017$ Day $1$, August $2, 2017$.

Let $f:\mathbb{R}\to(0,\infty)$ be a differentiable function, and suppose that there exists a constant $L>0$ such that $$|f'(x)-f'(y)|\leq L|x-y|$$ for all $x,y.$ Prove that $$(f'(x))^2<2Lf(x)$$ holds for all $x.$

The official solution goes as follows:

Solution: Notice that $f'$ satisfies the Lipschitz-property, so $f'$ is continuous and therefore locally integrable.

Consider an arbitrary $x\in\mathbb{R}$ and let $d=f'(x).$ We need to prove $f(x)>\frac{d^2}{2L}.$

If $d=0,$ then the statement is trivial.

If $d>0,$ then the condition provides $f'(x-t) \geq d-Lt;$ this estimate is positive for $1\leq t <\frac{d}{L}.$ By integrating over that interval, $$f(x)>f(x)-f(x-\frac{d}{L})=\int_0^{\frac{d}{L}}f'(x-t) \,dt \geq \int_0^{\frac{d}{L}}(d-Lt)\,dt=\frac{d^2}{2L}.$$ If $d<0,$ then apply $f'(x+t)\leq d+Lt=-|d|+Lt$ and repeat the same argument as $$f(x)>f(x)-f(x+\frac{|d|}{L}) = \int_0^{\frac{|d|}{L}}(-f'(x+t))\,dt \geq \int_0^{\frac{|d|}{L}}(|d|-Lt)\, dt = \frac{d^2}{2L}.$$

Question: What is a motivation behind the proof above?

Yes, I can fully understand the solution if I read line by line. However, I attempted the problem myself for 1 hour and could not get anywhere.

It would be good if someone can tell me a motivation of considering the proof above. It is okay if you want to provide an alternative solution to the problem. An alternative solution will give us another way of thinking the problem.

One idea could be to prove that the Taylor-like quadratic polynomial $$ g(s)=f(x)+f'(x)s+\frac{L}{2}s^2 $$ is an upper bound for $f(x+s)$. As $f$ is positive, so $g$ has to be positive, thus the minimum value of $g$, which by the vertex properties of quadratic functions is $$ g\left(-\frac{f'(x)}{L}\right)=\frac1{2L}\left(2Lf(x)-f'(x)^2\right), $$ has still to be positive.

By the mean value theorem, with some $θ\in(0,1)$, $$ f(x+s)-g(s)=(f'(x+\theta s)-g'(θs))s=(f′(x+θs)−f′(x)-Lθs)s. $$ As the difference of the derivatives is bounded by $Lθ|s|$, one gets $$ (f′(x+θs)−f′(x))s\le Lθs^2, $$ so that the whole expression is always non-positive, $$ f(x+s)\le g(s) $$

This is pretty much equivalent to the official solution, but from a somewhat different perspective. Maybe it will help you.

First, notice that both the given condition - let's label that as $(1)$ - and the one we want to prove - call that $(2)$ - are invariant under horizontal translations $x \rightarrow x+a$, so it suffices to prove $(f'(0))^2 < 2Lf(0)$ for any such function $f$. Then if we suppose that some $f$ satisfies $(1)$, but fails to meet $(2)$ somewhere else, say at $x = a$, just consider the function $g(x) = f(x+a)$; it also satisfies $(1)$ but fails to meet $(2)$ exactly at $x = 0$, which will be a contradiction once we prove this "restricted" version of $(2)$.

The approach to the proof is also by contradiction. Suppose that there exists some $f$ such that $(f'(0))^2 \geq 2Lf(0)$. Since $f(0) > 0$, this implies that either $f'(0) \geq \sqrt{2Lf(0)}$ or $f'(0) \leq -\sqrt{2Lf(0)}$.

Consider the first case: at $x = 0$, the function $f$ is increasing at a rate of $f'(0)$, and since $f'$ is continuous, the function must have been increasing over some interval to the left of $x = 0$. The idea is to show that the "net" increase over that interval would have to be greater than (or equal to) $f(0)$ which would be a contradiction because it would mean that $f$ must have been less than (or equal to) $0$ at the left endpoint.

So, a lower bound on the net increase, which is an integral of the derivative, will come from integrating a lower bound on the derivative. This we get from $(1)$ (applied to any non-positive $x$ and $y = 0$): $f'(x) \geq f'(0) + Lx$. We want this lower bound to be non-negative, so that we are certain $f$ is increasing, and that tells us to consider the interval $x \in \left[-{f'(0)\over L}, 0\right]$. Integrating yields $$ \int_{-f'(0)/L}^0 (f'(0)+Lx)dx = \frac{(f'(0))^2}{2L} \geq f(0) $$ by our assumption and the contradiction is achieved.

The case $f'(0) \leq -\sqrt{2Lf(0)}$ is analogous, just consider an interval to the right of $x = 0$ because now $f$ is decreasing at $x = 0$ and will be decreasing "for a while." We can show that it will have to "dip" at least as low as $0$ before the decrease can stop.