Dadgarnia wrote: $x,y$ and $z$ are real numbers such that $x+y+z=xy+yz+zx$. Prove that $$\frac{x}{\sqrt{x^4+x^2+1}}+\frac{y}{\sqrt{y^4+y^2+1}}+\frac{z}{\sqrt{z^4+z^2+1}}\geq \frac{-1}{\sqrt{3}}.$$ Proposed by Navid Safaei any solution.....

GGPiku wrote:

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Dadgarnia wrote: $x,y$ and $z$ are real numbers such that $x+y+z=xy+yz+zx$. Prove that $$\frac{x}{\sqrt{x^4+x^2+1}}+\frac{y}{\sqrt{y^4+y^2+1}}+\frac{z}{\sqrt{z^4+z^2+1}}\geq \frac{-1}{\sqrt{3}}.$$ Proposed by Navid Safaei The following is my proof. Maybe it's not very nice but I think it is the easiest and simpliest way WLOG, assume $x \ge y \ge z $ and denote $ f(x)=\frac{x}{\sqrt{x^4+x^2+1}}$. We see that $f$ increases on $[0, 1] $ , decreases on $[ 1, +\infty ]$ and $$ \max_{ x\ge 0} f(x) =f(1)=\frac{1}{\sqrt{3} }$$Case 1: $x \ge y \ge 0 \Longrightarrow LHS \ge -f(|z|) \ge -1/\sqrt{3}$ Case 2: $x \ge 0\ge y$, let $ b=-y, c=-z$ where $0 \le b \le c $. From $x+y+z=xy+yz+zx$ we get $x=\frac{b+c+bc}{b+c+1}$ Hence we need to prove $$ f(x)+f(1) \ge f(b)+f(c) $$If $f(x) \ge f(b)$ or $f(x) \ge f(c)$, we are done. Otherwise, $f(x) <f(b),f(c) $ , consider two small cases :

Case 3: $0 \ge x\ge y \Longrightarrow x+y+z \le 0 \le xy+yz+zx \Longrightarrow x=y=z=0 \Longrightarrow LHS =3>-1/\sqrt{3} $ Hence, the proof is completed.

Let $\displaystyle f(a)=\frac{a}{\sqrt{a^4+a^2+1}}$, notice that $$f'(a)=\frac{1-a^4}{(a^4+a^2+1)^{\frac{3}{2}}}$$Therefore, $f$ is increasing in the interval $[-1,1]$ and decreasing in the interval $[-\infty,-1]$ and $[1,\infty]$. Since $f(a)<0$ for all $a<0$ and $f(a)>0$ for all $a>0$, $f(-1)$ is the global minimum while $f(1)$ is the global maximum. Now notice that $x,y,z$ can not be all negative otherwise $x+y+z<0<xy+yz+zx$. Meanwhile if all $x,y,z$ are nonnegative then the statement is obvious. If one of $x,y,z$ is negative, say $z$ then $$f(x)+f(y)+f(z)\geq f(z)+0+0\geq f(-1)=\frac{-1}{\sqrt{3}}$$Now the only case left that two of $x,y,z$ say $x,y$ are negative. WLOG assume $|x|\geq|y|$. Then $$z+y=\frac{xy-x-y}{1-x-y}+y=\frac{-x-y^2}{1-x-y}$$We divide into three cases: Case I: $1\geq |x|$ Then $|y|\leq 1$, then $|x|\geq |y|\geq|y|^2$ so $-x-y^2\geq 0$, hence $z+y\geq0$, meanwhile $xy\leq 1$ so $z\leq 1$ as well, this implies $-f(z)=f(-z)\leq f(y)$ as $-z\leq y$, hence $$f(x)+f(y)+f(z)\geq f(x)\geq \frac{-1}{\sqrt{3}}$$Case II: $1\leq |y|$ We have $-y-x^2\leq 0$ so $z+x\leq 0$, therefore, $f(x)\geq f(-z)=-f(z)$ as $z\geq 1$. Hence $$f(x)+f(y)+f(z)\geq f(y)\geq\frac{-1}{\sqrt{3}}$$Case III: $|x|\geq 1\geq |y|$ This time $z+x\leq 0$ and $z+y\geq 0$. Hence if $z\geq 1$ then we finish similarly as in Case II, while if $z\leq 1$ we finish similarly as in Case I.