$\int_a^bf^2(x)\,dx\le \frac{2}{3}\int_a^bf(x)\,dx$ for a convex differentiable function If $f:[a,b] \to \mathbb{R}, f(a)=0,f(b)=1$  is a convex increasing differentiable function on the interval $[a,b]$ . Prove that 
$$\int_a^bf^2(x)\,dx\le \frac{2}{3}\int_a^bf(x)\,dx$$

Since f is convex and increasing so $f''(x)\ge 0 $ and $f'(x)\ge 0$. Then I consider a function $g:[a,b]\to \mathbb{R}$, $g(x)=\frac{2}{3}\int_a^xf(t)\,dt-\int_a^xf^2(t)\,dt$. Now $f$ is differentiable implies $g$ is also but can't conclude $g'(x)\ge 0$.
 A: In Prove $\int _0^\infty f^2 dx \leq \cdots $ for $f$ convex the following theorem was shown:

If $F$ is convex and non-negative on $[0, \infty)$ then $$
\int _0^\infty F^2(x) dx \leq \frac{2}{3}\cdot  \max_{x \in \mathbb R^+} F(x) \cdot  \int _0^\infty F(x) dx \, .$$

Our function $f$ is non-negative and convex on $[a, b]$ with $f(a) = 0$ and $f(b) = 1$. If we define $F$ on $[0, \infty)$ as
$$
 F(x) = \begin{cases}
f(b-x) & \text{ for } 0 \le x \le b-a \\
0 & \text{ for } x > b-a
\end{cases}
$$
then $F$ satisfies the hypotheses of the above theorem, and therefore
$$
\int_a^bf^2(x)\,dx = \int _0^\infty F^2(x) dx \leq \frac{2}{3}\cdot  \max_{x \in \mathbb R^+} F(x) \cdot  \int _0^\infty F(x) dx =  \frac{2}{3}\int_a^bf(x)\,dx \, .
$$

Alternatively we can modify the proof of the above theorem for this case. 
Define $\varphi: [a, b] \to \Bbb R$ as
$$
 \varphi(x) = \frac 23 f(x) \int_a^x f(t) \, dt - \int_a^x f^2(t) \, dt \, .
$$
The goal is to show that $\varphi$ is (weakly) increasing. Then the desired conclusion follows with
$$
 0 =  \varphi(a) \le  \varphi(b) = \frac 23  \int_a^b f(t) \, dt - \int_a^b f^2(t) \, .
$$
Since $f$ is assumed to be differentiable, we have
$$
 \varphi'(x) = \frac 23 f'(x) \int_a^x f(t) \, dt + \frac 23 f^2(x) - f^2(x) \\
 = \frac 23 f'(x) \int_a^x f(t) \, dt - \frac 13 f^2(x) \, .
$$
Now we distinguish two cases: 


*

*If $f'(x) =0$ then $f'(t) =0$ for $a \le t \le x$, so that $f(x) = f(a) = 0$  and therefore $\varphi'(x) = 0$.

*If $f'(x) >0$ then we estimate $f(t)$ from below by the tangent at $(x, f(x))$:
$$ 
\int_a^x f(t) \, dt \ge \int_{x-f(x)/f'(x)}^x \bigl( f(x) + (t-x)f'(x) \bigr) \, dt = \frac{f^2(x)}{2f'(x)}
$$
and therefore  $\varphi'(x) \ge 0$. 


So $\varphi'(x) \ge 0$ for all $x \in [a, b]$, which means that $\varphi$ is increasing on the interval, and we are done.
Remark 1: The proof becomes easier if we assume that $f$ is twice differentiable. Then
$$
 \varphi''(x) = \frac 23 f''(x)  \int_a^x f(t) \, dt \ge 0
$$
so that $\varphi'(x) \ge \varphi'(0) = 0$.
Remark 2: The proof works even without the assumption that $f$ is differentiable: As a convex function, $f$ has a right derivative
$$
f_+'(x) = \lim_{\substack{h \to 0\\ h > 0}} \frac{f(x+h)-f(x)}{h} 
$$
everywhere in $[a, b)$, and we can replace $f'$ by $f_+'$ and $\varphi'$ by $\varphi_+'$ in the above argument.
A: Without any loss of generality, we shift and scale to set $a=0, b=1$. And now we consider the integrals, $ \int_0^1{f(x)dx} $ and $ \int_0^1{f^2(x) dx} $. 
Convexity of $ f(x) $ assures that $f(x)\leq x$. (1)
Now, we write the integrals as limits of Riemann sums.$ \int_0^1{f^2(x) dx} = \lim_{h \to 0, N \to \infty}{\sum_{r=0}^N(f^2(rh) \times h)}$ $\int_0^1{f(x) dx} = \lim_{h \to 0, N \to \infty}{\sum_{r=0}^N(f(rh) \times h)} $
$ f^2(rh) / f(rh) =f(rh)  \leq rh$  (from (1)). For this ratio to be maximum (that is when the ratio $ \frac{ \int_0^1{f^2(x) dx}}{\int_0^1{f(x) dx}}$ is maximum), $ f(rh)=rh $ for all $r$. $\Rightarrow f(x)=x $. (2)
This means $ \frac{ \int_0^1{f^2(x) dx}}{\int_0^1{f(x) dx}} \leq \frac{ \int_0^1{x^2 dx}}{\int_0^1{x dx}} = \frac{2}{3}$
Edit: The maximisation holds if a unique maximum function exists in every interval of (2). This holds if $f(x)$ is convex. Otherwise as correctly pointed out by Martin, this ratio can be more than $\frac{2}{3}$. For example, if $ f(x)= sin^2(\frac{\pi x}{2})$, this ratio is 3/4. 
