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? 

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 $x=y=z=1$.
