Show $\int_0^t (t-x)P_n(x)\,dx\leq \frac{t^2}{2}\int_0^1 P_n(x)\,\mathrm dx $ where $P_n(x)=(x(1-x))^{n}$ 
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.
 A: Here is a proof using probabilistic arguments. 
$$f_n(t):=\frac{1}{k} P_n(t) \ \ \text{with} \ \ k:=\int_0^1 P_n(t)dt,$$ when restricted to interval $[0,1]$, is the pdf of a classical probability law : $\beta(n+1,n+1)$ (beta distribution).
Remark (to be used further on) : the curve of $f_n$ being symmetrical with respect to vertical line $t=1/2$, the curve of its cdf $F_n$ is symmetrical with respect to point $P(1/2,1/2)$ (see figure showing different cdf for $n=2\cdots 10$).
Dividing LHS and RHS of the given inequality 
$$t\int_0^t P_n(x)\,dx - \int_0^t xP_n(x)\,dx\leq \frac{t^2}{2}\int_0^1 P_n(x)\,dx\tag{*}$$
by the positive quantity $k$, it is equivalent to establish that
$$t\underbrace{\int_0^t f_n(x)dx}_{F_n(t)}-\int_0^t x f_n(x)dx \le t^2/2$$
$$\iff \ \ \forall t \in [0,1] : \ \ \underbrace{t^2/2 - tF_n(t) + \int_0^t xf_n(x)dx}_{\phi_n(t)} \geq 0 \tag{1}$$
Differentiating :
$$\phi_n'(t)=t-F_n(t)-\require{cancel} \cancel{tf_n(t)}+\cancel{tf_n(t)} \tag{2}$$
The curve of $y=t$ being symmetrical with respect to point $P(1/2,1/2)$, using the remark above, the curve of $\phi'$ will be symmetrical with respect to point $P$ ; therefore the curve of its primitive function will be symmetrical with respect to vertical line $t=1/2$. 
Then, it is sufficient to establish property (1) for $0 \leq t \leq 1/2$. Here is how. 
$F_n''(t)=f'_n(t)=kn(t(1-t))^{n-1}(1-2t)>0$ for $t \in (0,1/2)$ ; therefore, $F_n$ is convex in this domain ; consequently, as $F_n(0)=0$ and $F_n(1/2)=1/2$, the curve of $F_n$ is under the curve of $y=t$ for $t \in (0,1/2)$ ; 
We can conclude, using (2), that $\phi_n'(t)>0$ in $(0,1/2)$. As $\phi_n(0)=0$ we can infer that (1) is true, always in this interval $(0,1/2)$. Consequently, as said above, it is valid for the whole interval $[0,1]$.

A: 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.
