mberke wrote: Let $x,y,z$ be positive real numbers such that $x+y+z \ge x^2+y^2+z^2$. Show that; $$\dfrac{x^2+3}{x^3+1}+\dfrac{y^2+3}{y^3+1}+\dfrac{z^2+3}{z^3+1}\ge6$$ We need to prove that $\sum_{cyc}\left(\frac{x^2+3}{x^3+1}-2\right)\geq0$ or $\sum_{cyc}\frac{(1-x)(2x^2+x+1)}{x^3+1}\geq0$ or $\sum_{cyc}\left(2x(x-1)-\frac{(x-1)(2x^2+x+1)}{x^3+1}\right)+2\sum_{cyc}(x-x^2)\geq0$ or $\sum_{cyc}\frac{(x-1)^2(2x^3+2x^2+1)}{x^3+1}+2\sum_{cyc}(x-x^2)\geq0$. Done!

mberke wrote: Let $x,y,z$ be positive real numbers such that $x+y+z \ge x^2+y^2+z^2$. Show that; $$\dfrac{x^2+3}{x^3+1}+\dfrac{y^2+3}{y^3+1}+\dfrac{z^2+3}{z^3+1}\ge6$$

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mberke wrote: Let $x,y,z$ be positive real numbers such that $x+y+z \ge x^2+y^2+z^2$. Show that; $$\dfrac{x^2+3}{x^3+1}+\dfrac{y^2+3}{y^3+1}+\dfrac{z^2+3}{z^3+1}\ge6$$ $\sum {\frac{{x^2 + 3}}{{x^3 + 1}}} \ge \sum {\frac{{2x + 2}}{{\left( {x + 1} \right)\left( {x^2 - x + 1} \right)}}} = 2\sum {\frac{1}{{x^2 - x + 1}} \ge } 2 \times \frac{9}{{\sum {x^2 } - \sum x + 3}} \ge 6$

scpajmb wrote:

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$\sum {\frac{{x^2 + 3}}{{x^3 + 1}}} \ge \sum {\frac{{2x + 2}}{{\left( {x + 1} \right)\left( {x^2 - x + 1} \right)}} \ge } 2\sum {\frac{1}{{x^2 - x + 1}} \ge } 2 \times \frac{9}{{\sum {x^2 } - \sum x + 3}} \ge 6$ Good. $\sum \frac{x^2 + 3}{x^3 + 1}\ge \sum \frac{2x + 2}{x^3 + 1} = 2\sum \frac{1}{{x^2 - x + 1}} \ge \frac{2\cdot 9}{{\sum {x^2 } - \sum x + 3}} \ge 6.$

arqady wrote: mberke wrote: Let $x,y,z$ be positive real numbers such that $x+y+z \ge x^2+y^2+z^2$. Show that; $$\dfrac{x^2+3}{x^3+1}+\dfrac{y^2+3}{y^3+1}+\dfrac{z^2+3}{z^3+1}\ge6$$ We need to prove that $\sum_{cyc}\left(\frac{x^2+3}{x^3+1}-2\right)\geq0$ or $\sum_{cyc}\frac{(1-x)(2x^2+x+1)}{x^3+1}\geq0$ or $\sum_{cyc}\left(2x(x-1)-\frac{(x-1)(2x^2+x+1)}{x^3+1}\right)+2\sum_{cyc}(x-x^2)\geq0$ or $\sum_{cyc}\frac{(x-1)^2(2x^3+2x^2+1)}{x^3+1}+2\sum_{cyc}(x-x^2)\geq0$. Done! Sorry,don't understand...Could you please explain it..?

Let $x,y,z$ be positive real numbers such that $x+y+z \ge x^2+y^2+z^2$. Show that $$\sqrt{\frac{x^2+2}{2x+1}}+\sqrt{\frac{y^2+2}{2y+1}}+\sqrt{\frac{z^2+2}{2z+1}} \leq{\sqrt{\frac{x^3+2}{2x^2+1}}+\sqrt{\frac{y^3+2}{2y^2+1}}+\sqrt{\frac{z^3+2}{2z^2+1}}} $$

mberke wrote: Let $x,y,z$ be positive real numbers such that $x+y+z \ge x^2+y^2+z^2$. Show that; $$\dfrac{x^2+3}{x^3+1}+\dfrac{y^2+3}{y^3+1}+\dfrac{z^2+3}{z^3+1}\ge6$$ Let $x,y,z,m$ be positive real numbers such that $x+y+z \ge x^2+y^2+z^2$. Show that; $$(\frac{x^2+3}{x^3+1})^m+(\frac{y^2+3}{y^3+1})^m+(\frac{z^2+3}{z^3+1})^m\ge{3{\cdot}2^m} $$

Carl2015 wrote: Could you please explain it..? Yes, of course! What is your question?

I just wonder that how do you divide it into these different cases.Thx so much.

If I understood, your question is : what is a motivation to write $\sum_{cyc}\left(2x(x-1)-\frac{(x-1)(2x^2+x+1)}{x^3+1}\right)+2\sum_{cyc}(x-x^2)\geq0$ after $\sum_{cyc}\frac{(1-x)(2x^2+x+1)}{x^3+1}\geq0$? The equality in the starting inequality occurs for $x=y=z=1$. For $x=1$ the expression $\frac{2x^2+x+1}{x^3+1}$ equal to $2$. Hence, If we want to get $\lambda x(x-1)-\frac{(x-1)(2x^2+x+1)}{x^3+1}=(x-1)^2f(x)$, so we need to assume $\lambda=2$.

I see,thanks!My English is poor...

The original problem is a special case of the following generalization where $$n=3,k=2$$.

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Generalization 1 Let $x_{1},x_{2},\cdots,x_{n},k$ be positive reals ($k\geq 1$) such that $\sum_{cyc}{x_{1}}\geq \sum_{cyc}{x_{1}^2}$. Then prove that $$\sum_{cyc}{\dfrac{x_{1}^{k}+2k-1}{x_{1}^3+1}}\geq kn$$