TITLE: If $\int_0^x f^2(t)dt \le f(x)$ for all $x \in [0,1]$, then $\min_{[0,1]} f(x) \le 1$?
QUESTION [10 upvotes]: Suppose that $f$ is a continuous function on $[0,1]$ and 
$$\int_0^x [f(t)]^2dt \le f(x) \quad \text{for all} \quad x \in[0,1].$$
Prove or disprove
$$\min_{0\le x\le 1} f(x) \le 1.$$
In case the desired inequality does not hold, what is the best upper bound?
Thanks.

REPLY [7 votes]: We can show that
$$
 \min_{x\in[0,1]}\,f(x) < 2 \, ,
$$
and that is the best possible result, as Batominovski demonstrated with his counter-example.
Assume that $f(x) \ge 2$ in the interval. For $0 < x \le 1$ we define
$$
 g(x) = \frac{1}{\int_0^x f^2(t) \, dt}  \, .
$$
Then
$$
 g'(x) = -\frac{f^2(x)}{\left( \int_0^x f^2(t) \, dt\right)^2} \le -1
$$
and therefore
$$
 g(x) \ge g(1) + (1 - x) > 1 - x \, .
$$
This implies
$$
 4 x \le  \int_0^x f^2(t) \, dt = \frac{1}{g(x)} < \frac{1}{1-x}
$$
and setting $x = \frac 12$ gives a contradiction.