If $xy+yz+zx=3$ and $x,y,z\geq0$, prove that:
$$\sum_{cyc}\frac{1}{1+3x-p}\leq\frac{3}{1+2p}$$ where $p=xyz$.
*some people are not familiar with the $\sum_{cyc}$ notation, alternative would be
$$\frac{1}{1+3x-p}+\frac{1}{1+3y-p}+\frac{1}{1+3z-p}\leq\frac{3}{1+2p}$$
What I've tried:
- Using AM>GM, resulting $p\leq1$, if you use that directly it turns into a false inequality $\sum_{cyc}\cfrac{1}{3x}\leq1$
- Using the fact that $\frac{1}{1+3x-p}<\frac{1}{3x-p}$, resulting again in very clearly false inequality.
- Multipling by random stuff like $\sum_{cyc}(1+3x-p)$ and then applying the CBS inequality
- Deconditioning, first substituting $u=xy, v=yz, w=zx$ and then $u=\cfrac{3a}{a+b+c}, v=\cfrac{3b}{a+b+c}, w=\cfrac{3c}{a+b+c}$
What could be usefull: $$x+y+z\geq3$$ $$x^2+y^2+z^2\geq3$$ both prooven from $x^2+y^2+z^2\geq xy+yz+zx=3$, because of $(x-y)^2+(y-z)^2+(z-x)^2\geq0$
The inequality is equivalent to $$\sum_{cyc}\frac{3x-p}{1+3x-p}+\frac{3}{1+2p}\geq 3.$$ Notice that $$(3x-p)(y+z)=x(3-yz)(y+z)=x(xy+xz)(y+z)=(xy+xz)^2.$$ Therefore, by Cauchy-Schwarz, $$\sum_{cyc}\frac{3x-p}{1+3x-p}\cdot\sum_{cyc}(1+3x-p)(y+z) \geq 4(xy+yz+zy)^2=36$$ and it follows that $$\sum_{cyc}\frac{3x-p}{1+3x-p}\geq \frac{36}{\sum_{cyc}(1+3x-p)(y+z)}=\frac{18}{9+s(1-p)}$$ where $s=x+y+z$. Hence it suffices to show that $$\frac{18}{9+s(1-p)}+\frac{3}{1+2p}\geq 3$$ that is $$\frac{(1-p)(3-sp)}{(9+s(1-p))(1+2p)}\geq 0$$ which holds because $xy+yz+xz=3$ implies $p\leq 1$ and $sp\leq 3$.
P.S. $sp\leq 3$ iff $9=(xy+yz+zx)^{2} \geq 3xyz(x+y+z)$ iff $$2(xy+yz+zx)^{2}-6xyz(x+y+z)=x^2(y-z)^2+y^2(z-x)^2+z^2(x-y)^2\geq 0$$ which trivially holds.