# 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

$$x^n(1-x)+x(1-x)^n\leq \frac{x}{2n}+\frac{1-x}{2n}=\frac1{2n}.$$

• Why is $({n - 1 \over n})^{n-1}$ minimized when $n = 2$? Jan 29, 2020 at 5:39
• One can show that the function $f(x)=x\log(1+1/x)$ is increasing on $\mathbb R^+$. Jan 29, 2020 at 5:44
• I should have said maximized of course. So you're letting $x = n -1$ here and using that $(1 + 1/x)^x$ is increasing in $x$ essentially? Also at the end do you mean to use the same inequality, replacing $x$ by $1 - x$, and then adding the two inequalities? You didn't complete the argument. Jan 29, 2020 at 6:00
• @Zarrax Oops -- sorry about that. I've clarified now. Jan 29, 2020 at 6:18
• @CarlSchildkraut Thanks a lot for your input! This works. Jan 29, 2020 at 6:49