My solution Start by adding $3$ to both sides . Now we get $\sum (x+xy+y+1)/(xy+2y+1)\ge 3$. Now we have that that is equivavelnt to $\sum (x+1)(y+1)/(xy+2y+1) \ge 3$ Now by $CS$ we have $(\sum (x+1)(y+1)/(xy+2y+1)) (\sum (xy+2y+1)/(x+1)(y+1) ) \ge 9$ so now $LHS\ge 9/ (\sum (xy+2y+1)/(x+1)(y+1) )$ so now we get that inequality is equvalent to $3\ge \sum (xy+2y+1)/(x+1)(y+1) $ which is now equavalent to $3(x+1)(y+1)(z+1)\ge \sum (xy+2y+1)(z+1)=\sum xyz + 2yz+z+xy+2y+1=3xyz+3xy + 3xz+ 3yz +3x+3y+3z+3$ but we know that $3(x+1)(y+1)(z+1)=3xyz+3xy+3xz+3yz+3x+3y+3z+3$ so we see that inequality holds so we are done . Equality is trivial.

IgorM wrote: Let $x,y,z$ be nonnegative positive integers. Prove $\frac{x-y}{xy+2y+1}+\frac{y-z}{zy+2z+1}+\frac{z-x}{xz+2x+1}\ge 0$ $\frac{xy+2y+1}{(x+1)(y+1)}+\frac{yz+2z+1}{(y+1)(z+1)}+\frac{zx+2x+1}{(z+1)(x+1)}$ $=\frac{y(x+1)+y+1}{(x+1)(y+1)}+\frac{z(y+1)+z+1}{(y+1)(z+1)}+\frac{x(z+1)+x+1}{(z+1)(x+1)}=3$, $\frac{(x+1)(y+1)}{xy+2y+1}+\frac{(y+1)(z+1)}{yz+2z+1}+\frac{(z+1)(x+1)}{zx+2x+1}$ $\geq \frac{9}{\frac{xy+2y+1}{(x+1)(y+1)}+\frac{yz+2z+1}{(y+1)(z+1)}+\frac{zx+2x+1}{(z+1)(x+1)}}=3\implies$ $\frac{(x+1)(y+1)}{xy+2y+1}+\frac{(y+1)(z+1)}{yz+2z+1}+\frac{(z+1)(x+1)}{zx+2x+1}\geq 3\iff$ $\frac{x-y}{xy+2y+1}+\frac{y-z}{zy+2z+1}+\frac{z-x}{zx+2x+1}\ge 0$

The following inequality is also true. Let $x,y,z$ be nonnegative numbers.Prove that$$\frac{x-y}{xy+2y+3}+\frac{y-z}{zy+2z+3}+\frac{z-x}{xz+2x+3}\ge 0 .$$

hello, it is equivalent to $2 u^4+4 u^3 v+4 u^3+x^2 \left(2 u^2+2 u v+2 v^2\right)+2 u^2 v^2+11 u^2 v+6 u^2+x \left(4 u^3+6 u^2 v+4 u^2+2 u v^2+4 u v+4 v^2\right)+7 u v^2+6 u v+6 v^2\geq 0$ Sonnhard.

Dr Sonnhard Graubner, what about the case $x\geq y\geq z$?

If we add 3 to both sides we will get $\sum\frac{ (x+1)(y+1)}{(xy+2y+1)} \ge 3$ Then we can do a substution $x+1=a$ $\to$ $ x=a-1$ Then inequality get the form $\sum\frac{ ab}{ab+b-a} \ge 3$ The divide up and down by $ab$. Then equality get the form $\sum\frac{ 1}{1+\frac{1}{b}-\frac{1}{a}} \ge 3$ which is true by Cauchy Schwarz. Anar Abbas Istek Lyceum

The following inequality is also true. Let $x,y,z$ be nonnegative numbers.Prove that$$\frac{x-y}{xy+4y+3}+\frac{y-z}{zy+4z+3}+\frac{z-x}{xz+4x+3}\ge 0 .$$

sqing wrote: The following inequality is also true. Let $x,y,z$ be nonnegative numbers.Prove that$$\frac{x-y}{xy+4y+3}+\frac{y-z}{zy+4z+3}+\frac{z-x}{xz+4x+3}\ge 0 .$$ Id est we need to prove that $4\sum_{cyc}(x^2y^2-x^2yz)+\sum_{cyc}(19x^2z-3x^2y-16xyz)+8\sum_{cyc}(x^2-xy)\geq0$ or $2\sum_{cyc}z^2(x-y)^2+\sum_{cyc}(8x^2y+8x^2z-16xyz)+4\sum_{cyc}(x-y)^2\geq11\sum_{cyc}(x^2y-x^2z)$ or $\sum_{cyc}(x-y)^2(2z^2+8z+4)\geq11\sum_{cyc}(x^2y-x^2z)$, which is true because by AM-GM $\sum_{cyc}(x-y)^2(2z^2+8z+4)\geq(8+4\sqrt2)\sum_{cyc}(x-y)^2z\geq11\sum_{cyc}(x-y)^2z\geq11\sum_{cyc}(x^2y-x^2z)$.

The following inequality is also true. Let $x$, $y$ and $z$ be non-negative numbers. Prove that: $$\frac{x-y}{xy+y+1}+\frac{y-z}{yz+z+1}+\frac{z-x}{zx+x+1}\ge 0$$

Lemma: $\frac{x-y}{xy+2y+1}\ge \frac{x-y}{xy+x+y+1}$ Proof: Since $x,y\ge0$, it is equivalent to $(x-y)(xy+x+y+1)\ge (x-y)(xy+2y+1)$ $\iff (x-y)(x-y)\ge0 \iff (x-y)^2\ge0$ which is true. $\square$ Returning to our problem, we have $\sum_{cyc}\frac{x-y}{xy+2y+1}\ge \sum_{cyc}\frac{x-y}{xy+x+y+1}=\sum_{cyc}\frac{(x+1)-(y+1)}{(x+1)(y+1)}=\sum_{cyc}\frac{1}{y+1}-\frac{1}{x+1}=0$ $\Longrightarrow \sum_{cyc}\frac{x-y}{xy+2y+1}\ge0$, as desired. $\blacksquare$

sqing wrote: The following inequality is also true. Let $x,y,z$ be nonnegative numbers.Prove that$$\frac{x-y}{xy+4y+3}+\frac{y-z}{zy+4z+3}+\frac{z-x}{xz+4x+3}\ge 0 .$$

arqady wrote: The following inequality is also true. Let $x$, $y$ and $z$ be non-negative numbers. Prove that: $$\frac{x-y}{xy+y+1}+\frac{y-z}{yz+z+1}+\frac{z-x}{zx+x+1}\ge 0$$ Let $\text{min}(x,y,z)=z$. If $xy+y\ge x$, then it is $$(x-y)^2 (z^2(x+y)^2+1+z)+(xy+2y-x+1)(x-z)(y-z)\ge 0$$If $x\ge xy+y$, then it is $$(x-y)(x-xy-y+z(x+y-z))+(x-y)^2(z^2(x+y)^2+z)+(xy+1+y)(x-z)(y-z)\ge 0$$

arqady wrote: The following inequality is also true. Let $x$, $y$ and $z$ be non-negative numbers. Prove that: $$\frac{x-y}{xy+y+1}+\frac{y-z}{yz+z+1}+\frac{z-x}{zx+x+1}\ge 0$$ We have \[LHS=\frac{\sum{(xz-yz+x-z)^2}}{2(xy+y+1)(yz+z+1)(xz+x+1)}\ge{0}\]

IgorM wrote: Let $x,y,z$ be nonnegative positive integers. Prove $\frac{x-y}{xy+2y+1}+\frac{y-z}{zy+2z+1}+\frac{z-x}{xz+2x+1}\ge 0$ We have \[LHS=\frac{\sum{z(x-y)^2}+\sum{(xz-yz+x-z)^2}}{(xy+2y+1)(yz+2z+1)(xz+2x+1)}\ge{0}\]

sqing wrote: The following inequality is also true. Let $x,y,z$ be nonnegative numbers.Prove that$$\frac{x-y}{xy+2y+3}+\frac{y-z}{zy+2z+3}+\frac{z-x}{xz+2x+3}\ge 0 .$$ We have \[LHS=\frac{\sum{(z^2+12z+3)(x-y)^2}+5\sum{(xz-yz+x+y-2z)^2}}{6(xy+2y+3)(yz+2z+3)(xz+2x+3)}\ge{0}\]

sqing wrote: The following inequality is also true. Let $x,y,z$ be nonnegative numbers.Prove that$$\frac{x-y}{xy+4y+3}+\frac{y-z}{zy+4z+3}+\frac{z-x}{xz+4x+3}\ge 0 .$$ We have \[LHS=\frac{\sum{(xyz+2z^3+6xy+8xz+16z^2)(x-y)^2}}{2(xy+4y+3)(yz+4z+3)(xz+4x+3)(x+y+z)}+\frac{3\sum{(xy-yz+2y-2z)^2}}{2(xy+4y+3)(yz+4z+3)(xz+4x+3)}\ge{0}\]

szl6208 wrote: sqing wrote: The following inequality is also true. Let $x,y,z$ be nonnegative numbers.Prove that$$\frac{x-y}{xy+4y+3}+\frac{y-z}{zy+4z+3}+\frac{z-x}{xz+4x+3}\ge 0 .$$ We have \[LHS=\frac{\sum{(xyz+2z^3+6xy+8xz+16z^2)(x-y)^2}}{2(xy+4y+3)(yz+4z+3)(xz+4x+3)(x+y+z)}+\frac{3\sum{(xy-yz+2y-2z)^2}}{2(xy+4y+3)(yz+4z+3)(xz+4x+3)}\ge{0}\] how can you made that?!Im eager to hear that

IgorM wrote: Let $x,y,z$ be nonnegative positive integers. Prove $\frac{x-y}{xy+2y+1}+\frac{y-z}{zy+2z+1}+\frac{z-x}{xz+2x+1}\ge 0$ Let $x,y,z,k,m$ be nonnegative numbers,prove that $$\sum_{cyc}{\frac{x-y}{xy+(k+m)y+4k^2}}\ge 0$$