Inequality that holds for all $n\in \mathbb N$ and $x\in [0,1]$. I want to prove that 
$$x(1-x)^n+x^n(1-x)\le \frac1{2n}$$
for $x\in[0,1]$ and any $n\in \mathbb N$ ($n\ge1$). I have plotted the function $$f(x)=x(1-x)^n+x^n(1-x)-\frac1{2n}$$ and (if I am not mistaken) I know that it holds with equality for $n=1,2$ and strict inequality for $n\ge3$ but I cannot prove it rigorously. I have also tried to exploit symmetries, e.g., $g(x)=x(1-x)^n$, then I can write it as $g(x)+g(1-x)\le \frac{1}{2n}$ or maximize each summand independently (I know that $x^n(1-x)<1/ne$ with the max attained at $x=\frac{n}{n+1}$) or to write it as $$nx^{n-1}+n(1-x)^{n-1}\le \frac{1}{2x(1-x)}$$ and again exploit some kind of AM-GM inequality or $\ln$ type inequality (the RHS in the last inequality is $\left(\ln{\sqrt{\frac{x}{1-x}}}\right)'$, or do induction but nothing has worked. 
Edit: My best shot up to now is to write it as $$g(x)+g(1-x)\le0$$ with $g(x)=2nx^{n-1}-\frac1x$ and $x\in[0,1/2]$. The LHS is already less than $-1$ for $n\ge3$ so there seems to be enough leeway (at least more than in the original formulation). Also, the maximum of $g(x)+g(1-x)$ is attained at $1/2$ (which is convenient) for all $n\ge0$ except for $4\le n\le 14$.

We claim that
$$x^n(1-x)\leq \frac{x}{2n}.$$
This is equivalent to
$$x^{n-1}(1-x)\leq \frac{1}{2n}.$$
However, the maximum of the left side is attained when its derivative is $0$, as it is $0$ at both $0$ and $1$. This means
$$(n-1)x_0^{n-2}-nx_0^{n-1}=0\implies x_0=\frac{n-1}{n}.$$
So
$$x^{n-1}(1-x)\leq \left(\frac{n-1}{n}\right)^{n-1}\frac{1}{n}\leq \left(\frac12\right)^1\frac1n=\frac1{2n}.$$
Now we can finish by noting that
$$x^n(1-x)+x(1-x)^n\leq \frac{x}{2n}+\frac{1-x}{2n}=\frac1{2n}.$$
