<s>[INST] If $xy+yz+zx=3$ and $x,y,z\geq0$, prove that: $\frac{1}{1+3x-p}+\frac{1}{1+3y-p}+\frac{1}{1+3z-p}\leq\frac{3}{1+2p}$ 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 $(x-y)^2+(y-z)^2+(z-x)^2\geq0$
 [/INST] 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.
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