Prove that $\left|30240\int_{0}^{1}x(1-x)f(x)f'(x)dx\right|\le1$. 
Let $f\in C^{3}[0,1]$such that $f(0)=f'(0)=f(1)=0$ and $\big|f''' (x)\big|\le 1$.Prove that $$\left|30240\int_{0}^{1}x(1-x)f(x)f'(x)dx\right|\le1 .$$

I couldn't make much progress on this problem. I thought that maybe I should try using polynomial interpolation since I have a bound for $|f'''|$, but I can't determine the interpolation polynomial and I am quickly stuck (there is also the problem that I am dealing with both $f$ and $f'$ under the integral). Apart from this, I don't think that there is much we can do, the solution probably relies on this technique, but I can't make further progress.
 A: First, using integration by parts, and noting that $f(0)=f'(0)=0$ we get
$$
f(x)=\frac{x^2}{2}f''(0)+\frac{1}{2}\int_0^x(x-t)^2f^{(3)}(t)dt\tag{1}
$$
And from $f(1)=0$ we get also
$$
0=f(1)=\frac{1}{2}f''(0)+\frac{1}{2}\int_0^1(1-t)^2f^{(3)}(t)dt
$$
Hence
$$\eqalign{
0&=\frac{x^2}{2}f''(0)+
\frac{1}{2}\int_0^xx^2(1-t)^2f^{(3)}(t)dt\cr
&+
\frac{1}{2}\int_x^1x^2(1-t)^2f^{(3)}(t)dt
\tag{2}}
$$
Subtracting from $(1)$ from $(2)$ we get
$$f(x)=-\int_0^1k(x,t)f^{(3)}(t)dt\tag{3}$$ with
$$k(x,t)=\cases{\dfrac{x^2(1-t)^2-(x-t)^2}{2}& if $t\leq x$\cr
\dfrac{x^2(1-t)^2}{2}& if $ x\leq t$
}$$
(Note that $k(x,t)\ge 0$ for $0\leq x,t\leq 1$).
Now, we note that
$$\eqalign{I&=\int_0^1x(1-x)f(x)f'(x)dx=\int_0^1\left(x-\frac{1}{2}\right)(f(x))^2dx\\
&=\int_0^1\int_0^1\int_0^1\left(x-\frac{1}{2}\right)k(x,t)k(x,s)f^{(3)}(t)f^{(3)}(s)dtdsdx\\
&=\int_0^1\int_0^1H(t,s)f^{(3)}(t)f^{(3)}(s)dtds\tag{4}
}
$$
with
$$H(t,s)=\int_0^1\left(x-\frac{1}{2}\right)k(x,t)k(x,s)dx\tag{5}$$
Now, $H(s,t)=H(t,s)$ and for $0\leq t\leq s\leq 1$ we find
$$H(t,s)=\frac{1}{240} (s-1)^2 t \left(s^4 (t-2)+s^3 (t+2)+s^2 (1-4 t)+s t+(t-1) t^4\right)$$
It is not very easy but we can show (see the remark below,) that $H$ is nonnegative, and this is the crucial point. As it is depicted in the next figure

It follows from the condition $|f^{(3)}|\leq 1$ that
$$|{I}|\leq \int_0^1\int_0^1H(t,s)dtds=\frac{1}{30240}.$$
Remark. Let me add a proof of the positivity of $H(t,s)$. Indeed, It is a matter of verification that for $0<t\leq s\leq 1$ we have
$$\eqalign{\frac{240 }{(1-s)^2 t}H(t,s)&=
(1-t) \left(s^2-t^2\right)+2 s^2 (1-s) (s-t)\\&
+s (s+1) (1-s)^2 t+(1-t)^2 t^2 (t+1)}$$
which is  a sum of positive terms.
A: Just an idea put $f(x)=\frac{g(x^2(1-x)^2)}{x(1-x)}$ we have :
$$f'(x)=-\frac{(2x-1)(2(x-1)^2x^2g'((x-1)^2x^2)-g((x-1)^2x^2))}{x^2(1-x)^2}$$
So we have :
$$f'(x)f(x)x(1-x)=-\frac{(2x-1)(2(x-1)^2x^2g'((x-1)^2x^2)-g((x-1)^2x^2))}{x^2(1-x)^2}x(1-x)\frac{g(x^2(1-x)^2)}{x(1-x)}$$
Or :
$$f'(x)f(x)x(1-x)=-(2x-1)\Big(2g'((1-x)^2x^2)g((1-x)^2x^2)-\frac{g^2(x^2(1-x)^2)}{x^2(1-x)^2}\Big)$$
Now we can integrate by parts but I can't go further .
Update :
As I'm stuck with the substitution above I propose another one :
Put :
$$f(x)=g(\ln(x)-\ln(1-x))$$
Then :
$$f'(x)f(x)x(1-x)=g(\ln(x)-\ln(1-x))g'(\ln(x)-\ln(1-x))$$
Now we can make the substitution $y=\ln(x)-\ln(1-x)$ or $x=\frac{e^y}{e^y+1}$ in the integral . Another substitution and we see that there is a link with the Laplace transform .
Update 2:
It's really my last idea on this problem .
We have the condition :
$$|f'''(x)|\leq 1\quad (1)$$
but :
$$1=(x+(1-x))^n$$
So we can use binomials theorem to expand the terms .
Now we put :
$$f(x)=\operatorname{B}_{i,n}(x)$$
Where we speak about the Bernstein polynomials .
See here  (3.1) for the derivative of a Bernstein polynomial
Remains to compare the two quantities .
If it's really useless I shall delete .
Hope it inspire someone .
A: At first,
$$I=\int_0^1x(1-x)f(x)f'(x)\text{d}x \,\overset{IBP}{=\!=\!=}\,\frac12x(1-x)f^2(x)\bigg|_0^1+\frac12\int_0^1 (2x-1)f^2(x)\text{ d}x,$$
$$I=\frac12\int_0^1 (2x-1)f^2(x)\text{ d}x.\tag1$$
The given conditions mean that $f(x)$ has a root of multiplicity $2$
at $x=0$ and a simple root at $x=1.$
Taking in account the Besou theorem and assuming the condition $|f'''(x)| = 1,$ one can get
$$|f_{opt}(x)| = a(x-0)^2(1-x) = a(x^2-x^3),\quad a = \frac16,\tag2$$
$$|I|\le\dfrac1{72}\int_0^1(2x-1)(x^2-x^3)^2\text{ d}x = \dfrac1{30240}.$$
$\color{green}{\textbf{EDIT of 12.08.20.}}$
Let us integrate the given inequality
$$-1\le f'''(x) \le 1\tag{1n}$$
under the given conditions on the interval $(0,x),$ then
$$
\begin{cases}
-x\le f''(x) - f''(0)\le x\\[4pt]
-\frac12x^2 \le f'(x) - xf''\left(0\right) \le \frac12x^2\\[4pt]
-\frac16x^3 \le f(x) - \frac12x^2f''\left(0\right) \le \frac16x^3._{\Large\mathstrut}
\end{cases}\tag{2n}$$
From  $(2n.3)$ should
$$|6f(x)-3f''(0)x^2| \le x^3,\quad -1\le -3f''(0)\le1,$$
$$6|f(x)|\le |x^3+ax^2|,\quad |a|\le 1,\quad f(1)=0,$$
$$6|f(x)|\le x^2-x^3 = h(x).$$
This confirms formulas $(2).$
Similarly, from $(2n.2)$
$$|f'(x)| \le \frac12 x^2 -\frac13x = h'(x),$$
wherein $h'(x)$ is synchronized with $h(x).$
Therefore,
$$|I| \le \left|\int_0^1 x(1-x) h(x) h'(x) \text{ d}x\right| = \frac1{30240}.$$
A: $\color{green}{\textbf{Version of 15.08.20.}}$
At first,
$$I=\int_0^1x(1-x)f(x)f'(x)\text{d}x \,\overset{IBP}{=\!=\!=}\,\frac12x(1-x)f^2(x)\bigg|_0^1+\frac12\int_0^1 (2x-1)f^2(x)\text{ d}x,$$
$$I=\frac12\int_0^1 (2x-1)f^2(x)\text{ d}x.\tag1$$
Let
$$x = \frac{y+1}2,\quad \text{ d}x = \frac12\text{ d}y,\quad y = 2x-1,\quad g(y) = f\left(\frac{y+1}2\right), \tag2$$
then
$$f(x) = g(2x-1) = g(y),\quad f'''(x) = 8g'''(2x-1) = g'''(y),\tag3$$
$$I = \frac14\int\limits_{-1}^{1} y g^2(y)\text{ d}y,\tag4$$
under the conditions
$$g(-1) = g'(-1) = g(1) = 0,\quad |g'''(y)| \le \frac18.\tag5$$
Decomposition to the even and the odd parts
$$g(y)=g^\,_+(y)+g^\,_-(y),\quad g^\,_\pm(y) = \frac12(g(y)\pm g(-y)),\quad 
g^\,_\pm(-y) = \pm g^\,_\pm(y),\tag6$$
gives
$$I = \int\limits_{0}^{1} y g^\,_+(y)\,g^\,_-(y)\text{ d}y.\tag7$$
In accordance with the Shwartz inequality,
$$I^2 \le \int\limits_{0}^{1} \big(y g^\,_+(y)\big)^2\text{ d}y\cdot \int\limits_{0}^{1} g^2_-(y)\text{ d}y,\tag8$$
wherein  $(8)$ becames the equality if
$$|g^\,_-(y)| = y\,g^\,_+(y).$$
Then from $(5)$ should
$$g^\,_+(y) = (1-y^2)h(y),\quad \big|g^\,_-(y) \big| =(y-y^3)h(y)\tag{9}$$
Therefore, the function
$$g(y)=(1+y)(1-y^2) h(y)$$
maximizes $|I|$ under the conditions $(5)$ near  $y=\pm1.$
Taking in account the rest of the conditions, one can get $h(y) = \text{constant} =\frac1{48},$
$$g(y) = \frac1{48}(1+y)(1-y^2),\tag{10}$$
$$48^2I_{opt} = \int\limits_0^1 (y^3-y)^2\text{ d}y
= \int\limits_0^1 (y^6-2y^4+y^2)\text{ d}y  = \frac17-\frac25+\frac13 = \frac8{105},\tag{11}$$
$$\color{brown}{\mathbf{|I| \le \frac1{30240}}},$$
$$f_{opt}(x) = \pm g_{opt}(2x-1) = \pm \frac16 (x^2-x^3).$$
