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 $4\le n\le 14$.


1 Answer 1


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

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

  • $\begingroup$ Why is $({n - 1 \over n})^{n-1}$ minimized when $n = 2$? $\endgroup$
    – Zarrax
    Jan 29, 2020 at 5:39
  • $\begingroup$ One can show that the function $f(x)=x\log(1+1/x)$ is increasing on $\mathbb R^+$. $\endgroup$ Jan 29, 2020 at 5:44
  • $\begingroup$ 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. $\endgroup$
    – Zarrax
    Jan 29, 2020 at 6:00
  • $\begingroup$ @Zarrax Oops -- sorry about that. I've clarified now. $\endgroup$ Jan 29, 2020 at 6:18
  • $\begingroup$ @CarlSchildkraut Thanks a lot for your input! This works. $\endgroup$
    – Jimmy R.
    Jan 29, 2020 at 6:49

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