If $\int_0^x f^2(t)dt \le f(x)$ for all $x \in [0,1]$, then $\min_{[0,1]} f(x) \le 1$? 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.
 A: I assume that $f^2(t)$ means $\big(f(t)\big)^2$.  I have a very weak bound 
$$\min_{x\in[0,1]}\,f(x)<2\sqrt{2}\,,$$
and do not know how to improve it.  Maybe somebody can use my proof to get a better bound.
Suppose on the contrary that there exists a function $f:[0,1]\to\mathbb{R}$ satisfying
$$\int_0^x\,\big(f(t)\big)^2\,\text{d}t\leq f(x)\text{ for all }x\in[0,1]\tag{*}$$
such that
$$k:=\min_{x\in[0,1]}\,f(x)\geq 2\sqrt{2}\,.$$
From (*), we get that
$$f(x)\geq \max\big\{k^2x,k\}\text{ for all }x\in[0,1]\,.$$
We use (*) once again and find that
$$f(x)\geq \int_0^{\frac1k}\,k^2\,\text{d}x+\int_{\frac{1}{k}}^x\,(k^2t)^2\,\text{d}t=\frac{2k}{3}+\frac{k^4x^3}{3}$$
for $x\in\left[\dfrac1k,1\right]$.  
Define polynomials $P_0$, $P_1$, $P_2$, $\ldots$ as follows:
$$P_0(z):=1\,,\,\,P_1(z):=z\,,\,\,P_2(x):=\frac{2}{3}+\frac{z^3}{3}\,,$$
and
$$P_r(z)=1+\int_{1}^z\,\big(P_{r-1}(\zeta)\big)^2\,\text{d}\zeta$$
for $r=3,4,5,\ldots$.  It can be proven by induction that, for each $r=0,1,2,\ldots$, $$f(x)\geq k\,P_r(kx)$$ for all $x\in\left[\dfrac1k,1\right]$.
We can prove by induction that $0\leq P_r(z)\leq 1$ for every $z\in[0,1]$.  That is, the constant term of $P_r$ is nonnegative for every $r=0,1,2,\ldots$.  It is obvious that the coefficients of higher-order terms in $P_r$ are nonnegative.  Furthermore, the degree of $P_r$ is $2^r-1$, and the coefficient $\lambda_r$ of the $(2^r-1)$-st degree term of $P_r$ is given by the recurrence relation
$$\lambda_r=\frac{1}{2^r-1}\lambda_{r-1}^2\,.$$
In other words,
$$\lambda_r=\frac{1}{\prod\limits_{j=1}^r\,(2^j-1)^{2^{r-j}}}\geq \frac{1}{\prod\limits_{j=2}^r\,2^{j\cdot2^{r-j}}}=\frac{1}{2^{3\cdot 2^{r-1}-r-2}}\,.$$
That is,
$$f(1)\geq k\,P_r(k)\geq \frac{k^{2^r}}{2^{3\cdot2^{r-1}-r-2}}$$
for every $r=0,1,2,\ldots$.  However, as $k\geq 2\sqrt{2}$, $k^{2^r}$ grows faster than $2^{3\cdot2^{r-1}-r-2}$, namely,
$$\lim_{r\to\infty}\,\frac{k^{2^r}}{2^{3\cdot2^{r-1}-r-2}}=\infty\,.$$
This yields a contradiction. 
Remark.  I believe that the polynomials $P_r$ converge pointwise, as $r\to\infty$, to $P$ on $(-2,+2)$, where $$P(z):=\frac{1}{2-z}\text{ for }z\in(-2,+2)\,.$$
I conjecture also that, for $z\geq 2$, $\lim\limits_{r\to\infty}\,P_r(z)=\infty$. If this is true, then it follows immediately that $\min\limits_{x\in[0,1]}\,f(x)<2$.
Counterexample. Interestingly, let $k\in(1,2)$ and define
$$f(x):=\max\left\{k,\frac{k}{2-kx}\right\}\text{ for all }x\in[0,1]\,.$$
Then, we see that $\min\limits_{x\in[0,1]}\,f(x)=k$.  Furthermore,
$$f(x)=k\geq k^2x=\int_0^x\,k^2\,\text{d}t=\int_0^x\,\big(f(t)\big)^2\,\text{d}t$$
for $x\in\left[0,\dfrac1k\right]$.  For $x\in\left[\dfrac1k,1\right]$, we have
$$f(x)=\frac{k}{2-kx}=k+\int_{\frac1k}^x\,\left(\frac{k}{2-kt}\right)^2\,\text{d}t=\int_0^x\,\big(f(t)\big)^2\,\text{d}t\,.$$
This shows that, if $\min\limits_{x\in[0,1]}\,f(x)<c$ for every such function $f$, then $c\geq 2$.
A: 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. 
