Prove this conjecture inequality $x\cdot \frac{(1-x)^{k-1}}{(k+1)^{k-2}}+\frac{(1-2x)^k}{k^k}\le \frac{1}{(k+2)^{k-1}}$

Question

let $x\in (0,1)$, and $k$ be postive intgers,such $k\ge 2$,

I conjecture following inequality maybe hold? $$x\cdot \dfrac{(1-x)^{k-1}}{(k+1)^{k-2}}+\dfrac{(1-2x)^k}{k^k}\le \dfrac{1}{(k+2)^{k-1}}$$ creat by wang yong xi

This is my attempt

when $k=2$,then inequality can be written as $$x\cdot (1-x)+\dfrac{(1-2x)^2}{4}\le\dfrac{1}{4}$$ it is obviously true.

when $k=3$then inequality can be written as $$\dfrac{x(1-x)^2}{4}+\dfrac{(1-2x)^3}{27}\le\dfrac{1}{25}$$ or $$-\dfrac{(5x-1)^2(5x+8)}{2700}\le 0$$ it is clearly true.

when $k=4$ it's equivalent $$\dfrac{x(1-x)^3}{25}+\dfrac{(1-2x)^4}{256}-\dfrac{1}{216}\le 0$$ or $$\dfrac{(6x-1)^2(108x^2+12x-125)}{172800}\le 0$$ it is clearly

when $k=5$, it's equivalent $$x\cdot\dfrac{(1-x)^4}{216}+\dfrac{(1-2x)^5}{5^5}-\dfrac{1}{7^4}=-\dfrac{(7x-1)^2(185563x^3-181202x^2-127589x+156384)}{1620675000}<0$$ But I can't prove for any postive intgers $k$.and I have found $$LHS-RHS=[(k+2)x-1]^2\cdot h(x,k)$$.so we must prove $h(x,k)\le 0$ Thanks

Answer 1

Consider only $k>2$. Denote $$f(x)=x\cdot \frac{(1-x)^{k-1}}{(k+1)^{k-2}}+\frac{(1-2x)^k}{k^k},$$ we need to prove that $f(x)\leqslant f(\frac1{k+2})$ for $x\in [0,1]$. We have $f'(\frac1{k+2})=0$ and $$f'(0)=(k+1)^{2-k}-2k^{1-k}=k^{2-k}\left(\left(1+\frac1k\right)^{2-k}-\frac2k\right)>0$$ by Bernoulli inequality $(1+x)^a>1+ax$ for $a=2-k$, $x=1/k$. Also $f(1)$ equals $\pm f(0)$. It means that the maximal value of $f$ on $[0,1]$ is attained at an interior point $a\in (0,1)$, and thus $f'(a)=0$.

I claim that $f'$ has unique root (multiplicity counted) on $(0,1/2]$, and this root is $\frac1{k+2}$, and at most one root on $[1/2,1)$. In both cases the only possible maximum point is $\frac{1}{k+2}$ (the second extremal point would be a local minimum)

We have $$f'(x)=(1-x)^{k-2}(1-kx)(k+1)^{2-k}-2(1-2x)^{k-1}k^{1-k}=0.$$ Denote $y=\frac{1-x}{1-2x}$, then $x=\frac{1-y}{1-2y}$, $1-kx=\frac{(k-2)y-(k-1)}{1-2y}$. Our equation $f'(x)=0$ in terms of $y$ rewrites as $$y^{k-2}((k-2)y-(k-1))(k+1)^{2-k}+2k^{1-k}=0.$$ LHS is monotone in $y$ for $y\in (1,\infty)$ (corresponds to $x\in (0,1/2)$) and for $y<0$ (corr. $x\in (1/2,1)$), that implies the above claim.