Show $\int_0^t (t-x)P_n(x)\,dx\leq \frac{t^2}{2}\int_0^1 P_n(x)\,\mathrm dx $ where $P_n(x)=(x(1-x))^{n}$

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 $[0,1]$.

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.

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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.