Because $x_{i}^{3}-x_{i+1}^{3}=(x_{i}-x_{i+1})(x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2})$, we have \[\sum_{i=1}^{n}\frac{x_{i}^{3}}{x_{i}^{2}-x_{i}x_{i+1}+x_{i+1}^{2}}=\sum_{i=1}^{n}\frac{x_{i}^{3}-x_{i+1}^{3}+x_{i+1}^{3}}{x_{i}^{2}-x_{i}x_{i+1}+x_{i+1}^{2}}=\] \[=\sum_{i=1}^{n}\left(x_{i}-x_{i+1}+\frac{x_{i+1}^{3}}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}\right)=\sum_{i=1}^{n}\frac{x_{i+1}^{3}}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}.\] So the inequality to prove: \[\sum_{i=1}^{n}\frac{x_{i}^{3}}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}\geq\frac{1}{3}(x_{1}+\dots+x_{n}).\] is equivalent to \[{\sum_{i=1}^{n}\frac{x_{i}^{3}+x_{i+1}^{3}}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}}\geq\frac{2}{3}(x_{1}+\dots+x_{n}).(1)\] Now, it is natural to try ${\frac{x_{i}^{3}+x_{i+1}^{3}}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}}\geq\alpha(x_{i}+x_{i+1})$. Assuming such an inequality true, checking the equality in $(1)$, we see that $\alpha$ should be $\frac{1}{3}$. So it remains to establish the inequality: $\frac{x^{3}+y^{3}}{x^{2}+xy+y^{2}}\geq\frac{1}{3}(x+y)$. But this is equivalent to $3(x^{2}-xy+y^{2})\geq x^{2}+y^{2}+xy$, or $2(x-y)^{2}\geq0$.

N.T.TUAN wrote: If $x_{1}, x_{2}, . . . , x_{n}$ are positive numbers, prove the inequality $\frac{x_{1}^{3}}{x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}}+\frac{x_{2}^{3}}{x_{2}^{2}+x_{2}x_{3}+x_{3}^{2}}+...+\frac{x_{n}^{3}}{x_{n}^{2}+x_{n}x_{1}+x_{1}^{2}}\geq\frac{x_{1}+x_{2}+...+x_{n}}{3}$. My proof: for all positive $u$ and $v$ holds $\frac{u^{3}}{u^{2}+uv+v^{2}}\geq\frac{2u-v}{3}.$

Me too arqady wrote: N.T.TUAN wrote: If $x_{1}, x_{2}, . . . , x_{n}$ are positive numbers, prove the inequality $\frac{x_{1}^{3}}{x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}}+\frac{x_{2}^{3}}{x_{2}^{2}+x_{2}x_{3}+x_{3}^{2}}+...+\frac{x_{n}^{3}}{x_{n}^{2}+x_{n}x_{1}+x_{1}^{2}}\geq\frac{x_{1}+x_{2}+...+x_{n}}{3}$. My proof: for all positive $u$ and $v$ holds $\frac{u^{3}}{u^{2}+uv+v^{2}}\geq\frac{2u-v}{3}.$

arqady wrote: N.T.TUAN wrote: If $x_{1}, x_{2}, . . . , x_{n}$ are positive numbers, prove the inequality $\frac{x_{1}^{3}}{x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}}+\frac{x_{2}^{3}}{x_{2}^{2}+x_{2}x_{3}+x_{3}^{2}}+...+\frac{x_{n}^{3}}{x_{n}^{2}+x_{n}x_{1}+x_{1}^{2}}\geq\frac{x_{1}+x_{2}+...+x_{n}}{3}$. My proof: for all positive $u$ and $v$ holds $\frac{u^{3}}{u^{2}+uv+v^{2}}\geq\frac{2u-v}{3}.$ \[\frac{x^{3}}{x^{2}+xy+y^{2}}=x-\frac{yx^{2}+y^{2}x}{x^{2}+xy+y^{2}}\] Then using AM-GM \[\frac{yx^{2}+y^{2}x}{x^{2}+xy+y^{2}}\le{\frac{x+y}{3}}\] then using this inequality \[\sum{\frac{x^{3}}{x^{2}+xy+y^{2}}}=\sum{x-\frac{yx^{2}+y^{2}x}{x^{2}+xy+y^{2}}}\geq{\sum{\frac{{2x-y}}{3}}=\frac{\sum{x}}{3}}\]

If $x_{1}, x_{2}, . . . , x_{n}$ are positive numbers, for $\lambda\leq1$ ,prove the inequality $\frac{x_{1}^{3}+\lambda x_1x_2( x_1+x_2)}{x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}}+\frac{x_{2}^{3}+\lambda x_2x_3( x_2+x_3)}{x_{2}^{2}+x_{2}x_{3}+x_{3}^{2}}+...+\frac{x_{n}^{3}+\lambda x_nx_1( x_n+x_1)}{x_{n}^{2}+x_{n}x_{1}+x_{1}^{2}}\geq\frac{1+2\lambda}{3}(x_{1}+x_{2}+...+x_{n})$.

$\frac{a^2+b^2+(2\lambda-1)ab}{a^2+b^2+ab}\ge\frac{1+2\lambda}{3}\iff (3-1-2\lambda)(a^2+b^2)+(6\lambda-3-1-2\lambda)ab\ge 0\iff 2(1-\lambda)(a^2+b^2-2ab)\ge 0\iff (1-\lambda)(a-b)^2\ge 0$ is true for all $a,b\in R_+\wedge \lambda\le 1$. Equality iff $\lambda =1\vee a=b$ Now $\sum_{i=1}^n\frac{x_{i}^{3}+\lambda x_{i+1}x_i( x_i+x_{i+1})}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}-\sum_{i=1}^n\frac{x_{i+1}^{3}+\lambda x_{i+1}x_i( x_i+x_{i+1})}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}=\sum_{i=1}^n(x_i-x_{i+1})=0$ where $x_{n+1}=x_1$ Hence $2\sum_{i=1}^n\frac{x_{i}^{3}+\lambda x_{i+1}x_i( x_i+x_{i+1})}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}=\sum_{i=1}^n\left(\frac{x_{i}^{3}+\lambda x_{i+1}x_i( x_i+x_{i+1})}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}+\frac{x_{i+1}^{3}+\lambda x_{i+1}x_i( x_i+x_{i+1})}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}\right)=\sum_{i=1}^n\frac{( x_i+x_{i+1})(x_i^2+(2\lambda-1)x_ix_{i+1}+x_{i+1}^3)}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}\ge \sum_{i=1}^n\frac{1+2\lambda}{3}( x_i+x_{i+1})=2\cdot \frac{1+2\lambda}{3}\cdot \sum_{i=1}^n x_i$ QED equality iff $\lambda=1\vee x_1=x_2=...=x_n$

WolfusA wrote: $\frac{a^2+b^2+(2\lambda-1)ab}{a^2+b^2+ab}\ge\frac{1+2\lambda}{3}\iff (3-1-2\lambda)(a^2+b^2)+(6\lambda-3-1-2\lambda)ab\ge 0\iff 2(1-\lambda)(a^2+b^2-2ab)\ge 0\iff (1-\lambda)(a-b)^2\ge 0$ is true for all $a,b\in R_+\wedge \lambda\le 1$. Equality iff $\lambda =1\vee a=b$ Now $\sum_{i=1}^n\frac{x_{i}^{3}+\lambda x_{i+1}x_i( x_i+x_{i+1})}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}-\sum_{i=1}^n\frac{x_{i+1}^{3}+\lambda x_{i+1}x_i( x_i+x_{i+1})}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}=\sum_{i=1}^n(x_i-x_{i+1})=0$ where $x_{n+1}=x_1$ Hence $2\sum_{i=1}^n\frac{x_{i}^{3}+\lambda x_{i+1}x_i( x_i+x_{i+1})}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}=\sum_{i=1}^n\left(\frac{x_{i}^{3}+\lambda x_{i+1}x_i( x_i+x_{i+1})}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}+\frac{x_{i+1}^{3}+\lambda x_{i+1}x_i( x_i+x_{i+1})}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}\right)=\sum_{i=1}^n\frac{( x_i+x_{i+1})(x_i^2+(2\lambda-1)x_ix_{i+1}+x_{i+1}^3)}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}\ge \sum_{i=1}^n\frac{1+2\lambda}{3}( x_i+x_{i+1})=2\cdot \frac{1+2\lambda}{3}\cdot \sum_{i=1}^n x_i$ QED equality iff $\lambda=1\vee x_1=x_2=...=x_n$ Thanks.

$x,y,z>0$,prove \[\sqrt {{\frac {{x}^{3}}{2\,{y}^{2}-yz+2\,{z}^{2}}}}+\sqrt {{\frac {{y} ^{3}}{2\,{z}^{2}-xz+2\,{x}^{2}}}}+\sqrt {{\frac {{z}^{3}}{2\,{x}^{2}-x y+2\,{y}^{2}}}}\geq \sqrt {x+y+z} \]

N.T.TUAN wrote: If $x_{1}, x_{2}, . . . , x_{n}$ are positive numbers, prove the inequality $\frac{x_{1}^{3}}{x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}}+\frac{x_{2}^{3}}{x_{2}^{2}+x_{2}x_{3}+x_{3}^{2}}+...+\frac{x_{n}^{3}}{x_{n}^{2}+x_{n}x_{1}+x_{1}^{2}}\geq\frac{x_{1}+x_{2}+...+x_{n}}{3}$. Just apply Jensen's inequality

arqady wrote: N.T.TUAN wrote: If $x_{1}, x_{2}, . . . , x_{n}$ are positive numbers, prove the inequality $\frac{x_{1}^{3}}{x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}}+\frac{x_{2}^{3}}{x_{2}^{2}+x_{2}x_{3}+x_{3}^{2}}+...+\frac{x_{n}^{3}}{x_{n}^{2}+x_{n}x_{1}+x_{1}^{2}}\geq\frac{x_{1}+x_{2}+...+x_{n}}{3}$. My proof: for all positive $u$ and $v$ holds $\frac{u^{3}}{u^{2}+uv+v^{2}}\geq\frac{2u-v}{3}.$ Simple and wonderful.