Let $f:[1,+\infty)\to \mathbb{R}$ be a positive, continuous function. If $\int_1^xf(t)\,\mathrm{d}t\leqslant f^2(x)$ for all $x\geqslant 1$, prove that $f(x)\geqslant\frac{1}{2}\,(x-1).$
Attempt. Usual procedures, like working on the monotonicity of the function $$g(x):=f(x)-\frac{1}{2}\,(x-1)$$ (in order to get $g$ increasing and therefore $g(x)\geqslant g(1)=f(1)>0$ for $x\geqslant 1$), do not work.
Thanks for the help.
Let $h(t)=f(1+t)$. We want to show that if $h:\mathbb{R}^+\to\mathbb{R}^+$ is a continuous function such that $$ \forall x>0,\quad\int_{0}^{x}h(t)\,dt \leq h(x)^2 \tag{A}$$ then $h(x)\geq \frac{x}{2}$. Let $H(x)=\int_{0}^{x}h(t)\,dt$. We have that $H$ is a positive, $C^1$ and increasing function fulfilling $$ \frac{H'(x)}{\sqrt{H(x)}} \geq 1 \tag{B}$$ and by integrating both sides of $(B)$ on $(0,z)$ we get $$ 2\sqrt{H(z)} \geq z \quad \leftrightarrow\quad H(z)\geq\frac{z^2}{4}.\tag{C}$$ The tricky part is just to convert this bound for $H(z)$ into a bound for its derivative:
$$ h(x)\geq \sqrt{\int_{0}^{x}h(t)\,dt}\geq\sqrt{\int_{0}^{x}\sqrt{\int_{0}^{t}h(u)\,du}\,dt}=\sqrt{\int_{0}^{x}\sqrt{H(t)}\,dt}\geq\sqrt{\int_{0}^{x}\frac{t}{2}\,dt}=\frac{x}{2}.\tag{D} $$ As a faster alternative, we may just invoke $(B)$ once again.