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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 $h(x,k)\le 0$ Thanks

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

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

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 $h(x,k)\le 0$ Thanks

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.

and I have solve when $k=4,5$ it is hold,But I can't prove for any $k\ge 6$.Thanks

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