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
$$(x+y+z+1)^2=5(xy+yz+xz)+1$$

Lemma($\color{Green}{\text{Vieta's formula}}$):
Let $\alpha_1$ 
be the root of the quadratic polynomial equation 
$aY^2+bY+c=0$;
then we have: $\alpha_2= \color{Blue}{\dfrac{-b}{a}}-\alpha_1$. 
Proof: 
Only notice that 
$\alpha_1 + \alpha_2 = \dfrac{-b}{a}$. 


$ 
\color{Purple} 
{ 
\text{Let's to look at one of the} 
\ \ 
x, y, z 
\ \ 
\text{as the} 
} $ 
$\color{Red}{\text{variable}}$ 
$ 
\color{Purple} 
{  
\text  {and to look at the others as} 
} $ 
constants.  

For example let's to look at $\color{Red}{y}$ as an indeterminate, 
and to look at $x,z$ as constants;
as @user399601 has been done.  
$$(x+\color{Red}{y}+z)^2+2(x+\color{Red}{y}+z)=5(x\color{Red}{y}+\color{Red}{y}z+zx) 
\Longrightarrow 
\\ 
\Bigg[ 
\color{Red}{y^2} 
+ 
\big(2(x+z)\big) 
\color{Red}{y} 
+ 
(x+z)^2 
\Bigg] 
+ 
\Bigg[ 
2 
\color{Red}{y} 
+(x+z) 
\Bigg] = 
\Bigg[ 
5(x+z) 
\color{Red}{y} 
+ 5zx 
\Bigg] 
\Longrightarrow 
\\ 
\color{Red}{y^2} 
+ 
\Big( 
2(x+z) 
+ 
2 
-5(x+z)
\Big) 
\color{Red}{y} 
+ 
\Big( 
(x+z)^2 
+ 
(x+z) 
-5zx 
\Big) 
=0  
\ \ \ \ \ \ \ \ \ \ \ 
\Longrightarrow 
\\ 
\color{Red}{y^2} 
+ 
\Big( 
\color{Blue}{ 
2 
-3(x+z)
} 
\Big) 
\color{Red}{y} 
+ 
\Big( 
(x+z)^2 
+ 
(x+z) 
-5zx 
\Big) 
=0  
\ \ \ \ \ \ \ 
\color{Green}{\star\star\star\star} 
$$


Suppose that 
$\color{Red}{y}$ 
satisfies the polynomial equation 
$\color{Green}{\star\star\star\star}$ 
;
then by 
$\color{Green}{\text{Vieta's formula}}$ 
; 
we can see that 
: 
$ 
\Big( 
\color{Blue}{ 
3(x+z) 
-2 
} 
- \color{Red}{y} 
\Big) 
$ 
will satisfies 
$\color{Green}{\star\star\star\star}$ 
. 
So we proved that:

If 
$(x,\color{Red}{y},z)$ satisfies 
$\color{Green}{\star\star\star\star}$ ; 
then 
$ 
( 
x 
, 
\color{Blue}{ 
3(x+z) 
-2 
} 
- \color{Red}{y} 
, 
z 
) 
$ 
will satisfies 
$\color{Green}{\star\star\star\star}$ 
.  
[ 
More specially if we let $x=1$ we have the following:
If 
$(1,\color{Red}{y},z)$ satisfies 
$\color{Green}{\star\star\star\star}$ ; 
then 
$ 
( 
x 
, 
\color{Blue}{ 
3z 
+ 
3 
-2 
} 
- \color{Red}{y} 
, 
z 
) 
$ 
will satisfies 
$\color{Green}{\star\star\star\star}$ 
. 
] 




$ 
\color{Purple} 
{  \text  {This method is called}}$ 
$ 
\color{Green} 
{  \text  {vieta-jumping}}$ 
For more information you can see here: 
https://math.stackexchange.com/questions/tagged/vieta-jumping
https://en.wikipedia.org/wiki/Vieta_jumping#Constant_descent_Vieta_jumping
