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? 

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!
