How to prove this inequality with the following condition? Let $x$, $y$, $z$ be three positive real numbers satisfying
\begin{equation}
x + y +z + 1 =4xyz.\tag{1}
\end{equation}
Prove  that
\begin{equation}
xy + yz + zx \geqslant x + y + z.\tag{2}
\end{equation}
I don't know how to start? 
 A: Firstly, I'm going to relabel $x$,$y$ and $z$ to $\alpha$,$\beta$ and $\gamma$.
We then consider the cubic polynomial
$$ f(x) = ax^3 + bx^2 + cx + d$$
Supposing this has roots $\alpha$,$\beta$ and $\gamma$, by using the factor theorem, expanding the brackets and collecting together coeffecients we get the relationships:
$$ -\frac{d}{a} =  \alpha\beta\gamma $$
$$ \frac{c}{a} = \alpha\beta + \beta\gamma + \gamma\alpha$$
$$ - \frac{b}{a} = \alpha + \beta + \gamma $$
We then have a condition on the coeffecients of the polynomial to consider instead. The inequality then should hopefully come out when considering suffecient conditions for a cubic polynomial to have three real positive roots.
A: Let $F(x,y,z)=x+y+z+1-4xyz$ and $G(x,y,z)=xy+yz+zx-x-y-z$. Then we wish to minimize $G$ subject to $F=0$, and so can use lagrange multipliers. Then the following are zero at a critical point: 
$$F_xG_y-F_yG_x=(x-y)(2z-1)^2,$$
$$F_yG_z-F_zG_y=(y-z)(2x-1)^2,$$
$$F_zG_x-F_xG_z=(z-x)(2y-1)^2.$$
Here we cannot have $x=y=z=1/2$ and satisfy $F=0$, so at any critical point two of the variables are equal. By symmetry put $x=y=s$ and $z=t$; then $G=0$ gives $t=1/(2s-1)$. And in this case $F$ becomes $s^2-2s+1=(s-1)^2 \ge 0.$
Note that this technique applies to the case $x,y,z \ge 0$, since $F=0$ is inconsistent with any of $x,y,z$ being $0$. So the min of $G$ must occur at a critical point, and that min is zero as above, making the desired inequality true.
A: Here is a boring, uninspiring but elementary proof ^_^. By $(1)$ and A.M.$\ge$G.M., we get $xyz = (x+y+z+1)/4 \ge (xyz)^{1/4}$. Hence $xyz\ge1$. WLOG, suppose $0<x\le y\le z$. There are three cases. Case 1: $x\ge1$. Clearly $(2)$ holds. Case 2: $0<x<1\le y< z$. Then $(2)$ also holds because
$$1-xyz\le0\le(1-x)(1-y)(1-z) = 1 - (x+y+z) + (xy+yz+zx) - xyz.$$
Case 3: $0<x\le y<1<z$. Then
\begin{align*}
(2)&\Leftrightarrow z(x+y) + xy \ge x+y+z,\\
&\Leftrightarrow z(x+y-1) + xy - x-y\ge0,\\
&\Leftrightarrow z(x+y-1)-1 + (x-1)(y-1) \ge0.
\end{align*}
So it suffices to show that $z(x+y-1)-1\ge0$. Since $x,y,z$ are positive, we must have $4xy>1$, otherwise the LHS of $(1)$ will be strictly greater than the RHS. Now $(1)$ implies that $z=(x+y+1)/(4xy-1)$. Therefore
$$
z(x+y-1)-1
=\frac{(x+y+1)(x+y-1)-(4xy-1)}{4xy-1}
=\frac{(x-y)^2}{4xy-1} \ge0
$$
and we are done. From the details of above three cases, it can be shown that equality in $(2)$ holds iff $x=y=z=1$.
A: Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, the condition gives $3u+1=4w^3$, which does not depend on $v^2$.
We need to prove that $v^2\geq u$, for which it's enough to prove it for a minimal value of $v^2$.
In another hand $x$, $y$ and $z$ are positive roots of the following equation.
$$X^3-3uX^2+3v^2X-w^3=0$$ or
$$3v^2X=-X^3+3uX^2+w^3.$$
Let $f(X)=-X^3+3uX^2+w^3$.
Hence, the line $Y=3v^2X$ and graph of $f$ have three common points and $v^2$ gets a minimal value,
when the line $Y=3v^2X$ is a tangent line to the graph of $f$, 
which happens for equality case of two variables.
Let $y=x$.
Hence, the condition gives $z=\frac{1}{2x-1}$, where $x>\frac{1}{2}$ and we need to prove that
$$x^2+\frac{2x}{2x-1}\geq2x+\frac{1}{2x-1}$$ or
$$x^2-2x+\frac{2x-1}{2x-1}\geq0$$ or
$$(x-1)^2\geq0.$$
Done!
