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

8 events

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Jun 8, 2018 at 3:51	comment	added	Fedor Petrov		@functionsug if $x_0\geqslant 1/2$ is another root of $f'$, we have $f'<0$ on $[\frac1{k+2},x_0]$, thus it can not be a maximum point
Jun 3, 2018 at 16:06	comment	added	math110		why the maximum point is interior $(0.1/2]$? maybe is interior on $[1/2,1)$.because there is a root on $[1/2,1)?$ can you explain more detail? Thanks
Oct 25, 2017 at 14:27	vote	accept	math110
Oct 25, 2017 at 13:45	comment	added	math110		ah,Nice Thanks, I have understand,+1
Oct 25, 2017 at 13:35	comment	added	Fedor Petrov		it works for negative $a$ too (and for $a\in (0,1)$ with the opposite sign). the function $(1+x)^a$ is convex, and we compare it with the tangent at $x=0$.
Oct 25, 2017 at 13:32	comment	added	math110		I think the Bernoulli inequality $(1+x)^a>1+ax$, the $a$ must $a>1$. but $2-k<0$
Oct 25, 2017 at 13:04	history	edited	Fedor Petrov	CC BY-SA 3.0	added 8 characters in body
Oct 25, 2017 at 12:55	history	answered	Fedor Petrov	CC BY-SA 3.0