my solution = $4+4xyz=2+2(x^2+y^2+z^2)\ge (1+x)(1+y)(1+z)$ for this we have $(x-1)^2+(y-1)^2+(z-1)^2\ge 0\Rightarrow x^2+y^2+z^2+4\ge 2(x+y+z)+1$ and $2(x^2+y^2+z^2)\ge 2(xy+yz+zx)$ and $x^2+y^2+z^2=2xyz+1$ adding yields $4+4xyz=2+2(x^2+y^2+z^2)\ge (1+x)(1+y)(1+z)$ so we are done

YESMAths wrote: Let $x,y,z$ be real numbers such that $x^2+y^2+z^2-2xyz=1$. Prove that \[ (1+x)(1+y)(1+z)\le 4+4xyz. \] Expanding the expression we have to show that $1+(x+y+z)+(xy+yz+zx)+xyz\leq 4+4xyz,$ or equivalently $(x+y+z)+(xy+yz+zx)\leq 3+3xyz.$ Using $AM-GM$ inequality, $x^2+1\geq 2x$, $y^2+1\geq 2y$ and $z^2+1\geq 2z$ are true. Summing them we have $2xyz+1=x^2+y^2+z^2\geq 2(x+y+z)-3,$ or equivalently $xyz+2\geq x+y+z. (1)$ Using the well-known inequality $x^2+y^2+z^2\geq xy+yz+zx$ and the given equation, $2xyz+1\geq xy+yz+zx$ $(2)$ is true. Adding $(1)$ and $(2)$, we get $3xyz+3\geq (x+y+z)+(xy+yz+zx),$ as desired.

I posted my solution....but people gave me 3 downvotes.....I don't know why.....whatever,I don't wanna get dowmvotes....so I have removed my solution....

AM-GM gives 3(xyz)^2/3>=1+2xyz , so xyz>=1. Let p=x+y+z, q=xy+yz+zx, r=xyz, then p^2>=3q>=3, and r>=1. So the condition is p^2-2q=1+2r, and to prove 1+p+q+r<=4+4r, which rewrites to p+q<=3/2(p^2-2q+1), so we have 2p+8q<=3p^2+3. But this is because 8/3p^2>=8q, and 1/3p^2+3>=2p <=> 1/3(p-3)^2>=0, which is true.

navi_09220114 wrote: AM-GM gives 3(xyz)^2/3>=1+2xyz , so xyz>=1. how do you get $xyz\geq 1$ $?$

$(1+x)(1+y)(1+z)\le 4+4xyz$ is equivalent to $1+x+y+z+xy+xz+yz\le 4+4xyz$ Since $4xyz=2(x^2+y^2+z^2)-2$ we must prove that \[1+x+y+z+xy+xz+yz\le 2(x^2+y^2+z^2)+2\]$\to$ \[x+y+z+xy+xz+yz+xyz\le 2(x^2+y^2+z^2)+1\]substuting $xyz$ with $\frac{x^2+y^2+z^2-1}{2}$ we must prove that \[2(x+y+z+xy+xz+yz+xyz)+x^2+y^2+z^2-1\le 4(x^2+y^2+z^2)+2\]which is equialent to \[2(x+y+z+xy+xz+yz)\le 3(x^2+y^2+z^2)+3\]since $2(x^2+y^2+z^2)\geq 2(xy+xz+yz)$ we must prove that \[2(x+y+z)\le x^2+y^2+z^2+3\]Which is clearly $\text{AM-GM}$ Since $x^2+1\geq 2x$

YESMAths wrote: Let $x,y,z$ be real numbers such that $x^2+y^2+z^2-2xyz=1$. Prove that \[ (1+x)(1+y)(1+z)\le 4+4xyz. \] Let $x=a+1$, $y=b+1$ and $z=c+1$. Hence, the condition gives $2abc=a^2+b^2+c^2-2(ab+ac+bc)$ and we need to prove that $(a+2)(b+2)(c+2)\leq4+4(a+1)(b+1)(c+1)$ or $3abc+2(ab+ac+bc)\geq0$ or $\sum_{cyc}(3a^2-2ab)\geq0$, which is obvious.

YESMAths wrote: Let $x,y,z$ be real numbers such that $x^2+y^2+z^2-2xyz=1$. Prove that \[ (1+x)(1+y)(1+z)\le 4+4xyz. \] $(1+x)(1+y)(1+z)=1+x+y+z+yz+zx+xy+xyz$ $\leq1+\sqrt{3(x^2+y^2+z^2)}+x^2+y^2+z^2+xyz=2+\sqrt{3+6xyz}+3xyz$ $=-\frac{1}{6}(\sqrt{3+6xyz}-3)^2+ 4+4xyz\leq 4+4xyz .$

henderson wrote: navi_09220114 wrote: AM-GM gives 3(xyz)^2/3>=1+2xyz , so xyz>=1. how do you get $xyz\geq 1$ $?$ Opps, it should be in fact the opposite... But I think it wont affect my proof though, so let me rewrite it here. Let p=x+y+z, q=xy+yz+zx, r=xyz, then the condition is p^2-2q=1+2r, and to prove 1+p+q+r<=4+4r, which rewrites to p+q<=3/2(p^2-2q+1), so we have 2p+8q<=3p^2+3. But this is because 8/3p^2>=8q, and 1/3p^2+3>=2p <=> 1/3(p-3)^2>=0, which is true. Equality when p^2=3q, and p=3, which means that x=y=z=1.

Anyhow using the given condition and expanding the required inequality will reveal the solution.

YESMAths wrote: Let $x,y,z$ be real numbers such that $x^2+y^2+z^2-2xyz=1$. Prove that \[ (1+x)(1+y)(1+z)\le 4+4xyz. \] We have $xyz=\frac{x^2+y^2+z^2-1}{2}$ Then $(1+x)(1+y)(1+z)\le 4+4xyz \iff $ $3(x^2+y^2+z^2)-2(xy+yz+zx)-2(x+y+z)+3\ge 0 \iff$ $ (x-1)^2+(y-1)^2+(z-1)^2+(x-y)^2+(y-z)^2+(z-x)^2\ge 0$

YESMAths wrote: Let $x,y,z$ be real numbers such that $x^2+y^2+z^2-2xyz=1$. Prove that \[ (1+x)(1+y)(1+z)\le 4+4xyz. \] We have \[4+4xyz-(1+x)(1+y)(1+z)=\frac{1}{2}\sum{(x-1)^2}+\frac{1}{2}\sum{(x-y)^2}\ge{0}\]

I think this is simple

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Nice problem. Expanding, we want to show that $xyz+xy+yz+zx+x+y+z+1\leq 2(x^2+y^2+z^2)+2$. Because $x^2+y^2+z^2\geq xy+yz+zx,$ we just have to establish that $xyz+x+y+z+1\leq x^2+y^2+z^2+2$. Turning back into the other form of the condition, we want to show establish that $x+y+z\leq xyz+3,$ and turning back again, we get want to show that $2(x+y+z)\leq x^2+y^2+z^2+3$. This rearranges to $\sum_{cyc} (x-1)^2\geq 0,$ and everything is reversible so we are done.