Finding $\min \frac{\left[\int_{0}^{1}f(x)\mathrm dx\right]^2}{\int_{0}^{1}(f(x))^3\mathrm dx}$ given that $f(0)=0$ and $0\lt f'(x)\le 1$ Let $f$ be a function having continuous derivative on $[0,1]$ such that $0 &lt; f' \le 1$ and $f(0)=0$. Defined $$I := \frac{\left(\int_0^1 f(x)dx\right)^2}{\int_0^1 f^3(x)dx}$$
we must have that:
$$(A) \quad I \ge \frac{2}{3}\qquad (B) \quad I \ge \frac{1}{2}\qquad (C) \quad I \ge \frac{1}{3}\qquad (D) \quad I \ge 1$$

I have been stuck on this for quite some time, here is what I have tried. Since $0\lt f'(x)\le 1$, then taking the definite integral on both sides of the inequality we get that $0\lt f(1)\le 1$. Let $f(1)=k\in (0,1]$. then the curve $y=f(x)$ passes through the points $(0,0)$ and $(1,k)$ the join of which lies on the line $y=kx$. I am unable to proceed from this step.
Another alternate approach I have tried to use was the Cauchy-Schwarz inequality for integrals but to no avail. Perhaps the AM-GM inequality might be useful, since the function is non-negative in $[0,1]$ but I am not exactly sure how. Any hints or ideas are appreciated. Thanks.
See we have: $0 &lt; f'(x) \le 1$ and $f(0)=0$ $$\implies \int_0^x 0dx&lt;\int_0^x f'(x)dx\le \int_0^x 1dx$$ $$\implies 0&lt;f(x)&lt;x, x\in [0,1]$$.
We also know $x \le x^n, 0&lt;n&lt;1 $ when $x\in [0,1]$.
Let $f(x) =x$, in the numerator since this is a potential function (it satisfies the constraints given). If we make the denominator larger we know $I$ must be greater than the following
$$I &gt;\frac{\left(\int_0^1 xdx \right)^2}{\int_0^1 (x^n)^{3}dx}, 0&lt;n&lt;1 \implies I &gt; \dfrac{3n+1}{4}, 0&lt;n&lt;1$$
If $n= 2/3$ for example we have that $I&gt;0.75$. And we see as $n$ approaches $1$ we have $I$ approaching $1$. Hence all 4 options are correct.
The simplest trial function I thought of is $f(x) = kx$, with $0 \leq k \leq 1$.  The numerator integral becomes $\frac{1}{2}kx^2$ which evaluates to $\frac{k}{2}$ so the numerator is $\frac{k^2}{4}.$. Meanwhile the denominator integral gives $\frac{1}{4}k^3x^4$ which evaluates to $\frac{1}{4}k^3$.  Dividing, I get the expression equals $k^{-1}$.  Since $k &lt; 1$, $1/k &gt; 1$, which does not rule out any of the options.
Next I try the trial function $f(x) = kx^p$.  For this we must have $kpx^{p-1} \leq 1$ for $0 \leq x \leq 1$.  Since $f(x)$ must be increasing, $p &gt; 0$ and $k &gt; 0$.  Also, if $p &lt; 1$ then the derivative at $0$ would be infinite, so we have $p \geq 1$, and we already tried $p = 1$.  So for given $k$ and $p$, $kpx^{p-1} \leq kp \leq 1$.  So $k \leq 1$.
The numerator integral is $\frac{k}{p+1}$ so the numerator is $\frac{k^2}{(p+1)^2}$, while the denominator integral is $\frac{1}{3p+1}k^3x^{3p+1}$ which evaluates to $\frac{k^3}{3p+1}$.  The expression becomes $$\frac{k^2(3p+1)}{(p+1)^2k^3} = \frac{3p+1}{k{(p+1)}^2}$$ Given that $k \leq 1$, we examine $\frac{3p+1}{{(p+1)}^2}$ which has an extremum at $p = \frac{1}{3}$. That looks promising, but plugging in, I get the expression is greater than $9/8$.
I begin to suspect the answer is $1$, but I am still trying to think of other functions that could somehow evaluate to less.  The target now becomes, prove $$\big[\int_0^1 f(x) dx \big]^2 \geq \int_0^1 (f(x))^3 dx$$