This is a question that I came across today:  

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


>I  conjecture following inequality maybe hold?
$$x\cdot \dfrac{(1-x)^{k-1}}{(k+1)^{k-2}}+\dfrac{(1-2x)^k}{k^k}\le \dfrac{1}{(k+2)^{k-1}}$$
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 $h(x,k)\ge 0$  Thanks