<s>[INST] How prove infinitely many postive integers triples $(x,y,z)$ such $(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$ 
show that there exsit infinitely many  postive integers triples $(x,y,z)$
such
$$(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$$

May try it is clear $(x,y,z)=(1,1,1)$ is one  solution，and
$$(x+y+z+1)^2=5(xy+yz+xz)+1$$
 [/INST] Here's a way to more constructively get the answer user399601 provided:
Note that since $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz)$, the equation 
\begin{equation}\label{eqn:constraint}\tag{1}
(x + y + z)^2 + 2(x + y + z) = 5(xy + yz + xz)
\end{equation}
is equivalent to
\begin{align*}
x^2 + y^2 + z^2 + 2(x + y + z) &= (x + 1)^2 + (y + 1)^2 + (z + 1)^2 - 3\\
&= 3xy + 3yz + 3xz
\end{align*}
Now take $z = 1$ to get
\begin{equation*}
(x + 1)^2 + (y + 1)^3 + 1 = 3(x + 1)(y + 1)
\end{equation*}
Making the substitutions $u = x+1$ and $v = y + 1$, the equation becomes
\begin{equation}\label{eqn:simplified}\tag{2}
u^2 + v^2 + 1 = 3uv
\end{equation}
and
\begin{equation*}
(u - v)^2 + 1 = uv
\end{equation*}
Now suppose there's some function $c:\mathbb{Z}^2 \to \mathbb{Z}$ such that $(u + c, v)$ is a solution whenever $(u, v)$ is a solution.  Then
\begin{align*}
(u - v + c)^2 + 1 &= (u + c)v\\
(u - v)^2 + 2c(u-v) + c^2 + 1 &= uv + cv \\
c^2 + 2(u - v)c &= cv
\end{align*}
So $$c(u, v) = 2(v - u) + v = 3v - 2u$$
Now note that if $(u, v)$ solves equation \eqref{eqn:simplified}, then $(u-1, v-1, 1)$ solves equation \eqref{eqn:constraint}.  Thus the solution $(3v - u, v)$ maps to $(3y - x + 1, y, 1)$.
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