So from your proof, equality when \[\frac{x^{k+2}}{x^{k+1}+2x^{k}}=\frac{1}{21}\] at $x=\frac{1}{3}$ so \[=\frac{x^{k+2}}{\frac{7}{3}x^{k}}=\frac{1}{21}\] which is an identity is for all $k$, there is equality at $x=y=z=\frac{1}{3}$

boy am i an idiot... just glad i never forgot to finish a solution in a competition thanks...

Just to give an example of your proof, can you show the steps when $k=1$?

not really sure what you mean because the steps are identical... DON'T look at this unless you wan't a really big hint... ok, the inequality for k=1 is this $\sum_{cyc}\frac{x^{3}}{x^{2}+y+z}\geq \frac{1}{7}$ where $x+y+z=1$

N.T.TUAN wrote: Let $k$ be a given natural number. Prove that for any positive numbers $x; y; z$ with the sum $1$ the following inequality holds: \[\frac{x^{k+2}}{x^{k+1}+y^{k}+z^{k}}+\frac{y^{k+2}}{y^{k+1}+z^{k}+x^{k}}+\frac{z^{k+2}}{z^{k+1}+x^{k}+y^{k}}\geq \frac{1}{7}.\]When does equality occur? See here: http://www.artofproblemsolving.com/community/c6h1142219p5392189

There's a wonderful solution in the book An Introduction to the Proving of Elementary Inequalities written by Jingjun Han.

Here is a bit of a different solution First of all notice that $\sum_{cyc}\frac{x^{k+2}}{x^{k+1}+y^k+z^k}\overset{\text{Generalized T2'S}}{\ge}\frac{\left(\sum_{cyc}x\right)^{k+2}}{3^k\left(\sum_{cyc}x^{k+1}+2\sum_{cyc}x^k\right)}$ Thus the inequality transforms into $7\left(\sum_{cyc}x\right)^{k+2}\ge3^k\left(\sum_{cyc}x^{k+1}+2\sum_{cyc}x^k\right)\Longleftrightarrow\frac{7}{3^k}\left(\sum_{cyc}x\right)^{k+2}\ge\sum_{cyc}x^{k+1}+2\sum_{cyc}x^k$ Moreover notice that from the problem condition we can furthermore rewrite the inequality as $\frac{7}{3^k}\ge\sum_{cyc}x^{k+1}+2\sum_{cyc}x^k$ Now let $f(x,y,z)=x^{k+1}+y^{k+1}+z^{k+1}+2x^k+2y^k+2z^k\text{ and }g(x,y,z)=x+y+z-1$ and also define $L=f(x,y,z)+\lambda g(x,y,z)$ $$\frac{\partial L}{\partial x}=kx^k+x^k+2kx^{k-1}+\lambda=0\Longrightarrow kx^k+x^k+2kx^{k-1}=-\lambda\color{red}\text{ (1) }$$$$\frac{\partial L}{\partial y}=ky^k+y^k+2ky^{k-1}+\lambda=0\Longrightarrow ky^k+y^k+2ky^{k-1}=-\lambda\color{blue}\text{ (2) }$$$$\frac{\partial L}{\partial z}=kz^k+z^k+2kz^{k-1}+\lambda=0\Longrightarrow kz^k+z^k+2kz^{k-1}=-\lambda\color{green}\text{ (3) }$$Furthermore from $\color{red}\text{ (1) }$ and $\color{blue}\text{ (2) }$ we obtain $kx^k+x^k+2kx^{k-1}=ky^k+y^k+2ky^{k-1}\Longrightarrow k(x^k-y^k)+2k(x^{k-1}-y^{k-1})=-(x^k-y^k)$ Now $\text{FTSOC}$ assume that $x\neq y$ With this assumption we can get $$k+\frac{2k(x^{k-1}+y^{k-1})}{x^k-y^k}=-1\Longrightarrow 1+\frac{2(x^{k-1}-y^{k-1})}{x^k-y^k}=-\frac{1}{k}\Longrightarrow\frac{2(x^{k-1}-y^{k-1})}{x^k-y^k}=-\frac{1}{k}-1\text{ we can furthermore simplify it to }\frac{x^{k-1}-y^{k-1}}{x^k-y^k}=-\frac{1+k}{2k}$$Thus $\frac{x^{k-1}-y^{k-1}}{x^k-y^k}<0$ which implies that either $x^{k-1}<y^{k-1}\text{ or }x^k<y^k$ Case 1: $x^{k-1}<y^{k-1}$ From this we obtain $x<y\Longrightarrow x^k<y^k$ which forces $\frac{\text{negative}}{\text{negative}}=\text{negative}$ which is clearly a contradiction. Case 2: $x^k<y^k$ From this we obtain $x<y\Longrightarrow x^{k-1}<y^{k-1}$ which furthermore forces $\frac{\text{negative}}{\text{negative}}=\text{negative}$ which is a contradiction. Therefore $x$ must be equal to $y$ Moreover from $\color{blue}\text{ (2) }$ and $\color{green}\text{ (3) }$ we obtain $ky^k+y^k+2ky^{k-1}=kz^k+z^k+2kz^{k-1}\Longrightarrow k(y^k-z^k)+2k(y^{k-1}-z^{k-1})=-(y^k-z^k)$ Now $\text{FTSOC}$ assume that $y\neq z$ Using this assumption we can rewrite the equation as $$k+\frac{2k(y^{k-1}-z^{k-1})}{y^k-z^k}=-1\Longrightarrow1+\frac{2(y^{k-1}-z^{k-1})}{y^k-z^k}=-\frac{1}{k}\Longrightarrow\frac{2(y^{k-1}-z^{k-1})}{y^k-z^k}=-\frac{1}{k}-1\text{ continuing on we can simplify it down to }\frac{y^{k-1}-z^{k-1}}{y^k-z^k}=-\frac{1+k}{2k}$$Thus $\frac{y^{k-1}-z^{k-1}}{y^k-z^k}<0$ which forces $y^{k-1}<z^{k-1}\text{ or }y^k<z^k$ Case 1: $y^{k-1}<z^{k-1}$ From this we can obtain $y<z\Longrightarrow y^k<z^k$ which implies that $\frac{\text{negative}}{\text{negative}}=\text{negative}$ which is clearly a contradiction. Case 2: $y^k<z^k$ Here we obtain $y<z\Longrightarrow y{k-1}<z^{k-1}$ which forces $\frac{\text{negative}}{\text{negative}}=\text{negative}$ which is a contradiction. Therefore $y$ must be equal to $z$ which in turn implies that $x=y=z$ Plugging this in we obtain $g(x,x,x)=3x-1=0\Longrightarrow x=y=z=\frac{1}{3}$ Therefore maximum is attained at $f\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)=3\cdot\frac{1}{3^{k+1}}+2\cdot3\cdot\frac{1}{3^k}=\frac{1}{3^k}+\frac{2}{3^{k-1}}=\frac{7}{3^k}$ $\blacksquare$.