TITLE: Show $\int_0^t (t-x)P_n(x)\,dx\leq \frac{t^2}{2}\int_0^1 P_n(x)\,\mathrm dx $ where $P_n(x)=(x(1-x))^{n}$
QUESTION [9 upvotes]: Show that for all $t\in [0,1]$, and for any $n\in\mathbb{N}$,
  $$\int_0^t (t-x)P_n(x)\,dx\leq \frac{t^2}{2}\int_0^1 P_n(x)\,dx\tag{*}$$
  where $P_n(x)=(x(1-x))^{n}$.

Since $P_n\geq  0$ over $[0,1]$ then  $\int_0^y P_n(x)\,dx\leq \int_0^1 P_n(x)\,dx$ for any $y\in [0,1]$, and it follows easily that for all $t\in [0,1]$,
$$\int_0^t (t-x)P_n(x)\,dx=\int_0^t\int_0^y P_n(x)\,dx\,dy\leq t\int_0^1 P_n(x)\,dx.$$
On the other hand, 
$$\int_0^t (t-x)P_n(x)\,dx\leq\int_0^t (t-x)\,dx \cdot \max_{x\in[0,1]}P_n(x)=\frac{t^2}{2}\max_{x\in[0,1]}P_n(x),$$
but $\max_{x\in[0,1]}P_n(x)>\int_0^1 P_n(x)\,dx$ for $n>0$.
The inequality with $(*)$ seems to be much harder. Is it known? Any reference is welcome.

REPLY [4 votes]: Let
$$f(t) =  \frac{t^2}{2}\int_0^1 x^n(1-x)^n \mathrm{d} x - \int_0^t (t-x)x^n(1-x)^n \mathrm{d} x.$$
It is easy to prove that $f(t) = f(1-t)$ for all $t$ in $[0, 1]$ (the proof is given at the end). 
Also, $f(0)=0$. Thus, it suffices to prove that $f(t) \ge 0$ for all $t$ in $(0, \frac{1}{2}]$.
We have
$$f'(t) = t\int_0^1 x^n(1-x)^n \mathrm{d} x - \int_0^t x^n(1-x)^n \mathrm{d} x.$$
Let
$$g(t) = \frac{\int_0^t x^n(1-x)^n \mathrm{d} x}{t}.$$
We have, for all $t$ in $(0, \frac{1}{2}]$,
\begin{align}
g'(t) &= \frac{t t^n(1-t)^n - \int_0^t x^n(1-x)^n \mathrm{d} x}{t^2}\\
&\ge \frac{t t^n(1-t)^n - \int_0^t t^n(1-t)^n \mathrm{d} x}{t^2}\\
&= 0
\end{align}
where we have used the fact that $x\mapsto x(1-x)$ is non-decreasing on $(0, \frac{1}{2}]$.
Thus, we have, for all $t$ in $(0, \frac{1}{2}]$,
\begin{align}
g(t) &\le g(\tfrac{1}{2})\\
& = 2\int_0^{1/2} x^n(1-x)^n \mathrm{d} x\\
&= \int_0^{1/2} x^n(1-x)^n \mathrm{d} x + \int_{1/2}^1 x^n(1-x)^n \mathrm{d} x\\
&= \int_0^1 x^n(1-x)^n \mathrm{d} x.
\end{align}
Thus, we have $f'(t) \ge 0$ for all $t$ in $(0, \frac{1}{2}]$. Thus, $f(t) \ge 0$ for all $t$ in $(0, \frac{1}{2}]$.
We are done.
$\phantom{2}$
Proof of $f(t)=f(1-t)$: Indeed, we have
\begin{align}
f(t) - f(1-t) &= \frac{t^2-(1-t)^2}{2}\int_0^1 x^n(1-x)^n \mathrm{d} x - \int_0^t (t-x)x^n(1-x)^n \mathrm{d} x\\
&\quad  + \int_0^{1-t} (1-t-x)x^n(1-x)^n \mathrm{d} x\\
&= \frac{2t-1}{2} \int_0^1 x^n(1-x)^n \mathrm{d} x - \int_0^t (t-x)x^n(1-x)^n \mathrm{d} x \\
&\quad + \int_t^1 (x-t)x^n(1-x)^n \mathrm{d} x \\
&= \frac{2t-1}{2} \int_0^1 x^n(1-x)^n \mathrm{d} x - \int_0^1 (t-x)x^n(1-x)^n \mathrm{d} x\\
&= -\frac{1}{2}\int_0^1 x^n(1-x)^n \mathrm{d} x  + \int_0^1 xx^n(1-x)^n \mathrm{d} x\\
&= -\frac{1}{2}\int_0^1 x^n(1-x)^n \mathrm{d} x\\
&\quad  + \frac{1}{2}
\left(\int_0^1 xx^n(1-x)^n \mathrm{d} x + \int_0^1 (1 - x)x^n(1-x)^n \mathrm{d} x\right)\\
&= 0.
\end{align}
We are done.