Are they equal

Let $\text{gcd}(x,y,z)=d$ . Then $x=ad, y=bd, z=cd$ , where $\text{gcd}(a,b,c)=1$ . By dividing with $d^2$ we get: $a(a+b)=b(b+c)=c(c+a)$ Since $a^2+ab=b^2+bc$ , $b\mid a^2$ . Similarly $c \mid b^2$ and $a\mid c^2$ . If $a>1$ , then let $p$ a prime divisor of $a$ . Then $p\mid c^2\Rightarrow p\mid c$ , but $p\mid b^2\Rightarrow p\mid b$ . However it contradicts that $\text{gcd}(a,b,c)=1$ . So $a=b=c=1$ , and from that $x=y=z$ is definitely true.

Remembering x,y,z are all positive

x(x+y) = y(y+z) gives

x^2 - y^2 = y(z-x) eq1

Sim: y(y+z) = z(z+x) gives y^2 - z^2 = z(x-y) eq2

If x > y then eq1&2 are positive on both sides, so z>x and y>z

Giving x>y>z>x which cannot be true

Sim: x<y gives x<y<z<x which cannot be true

So x is neither more nor less than y, they must be equal. Which means z is also equal. (From original eqs).

Divide by xyz:

$\frac{x+y}{yz}$ = $\frac{y+z}{xz}$ = $\frac{z+x}{xy}$

and, by the property that a/b=c/d means they're also equal to (a-c)/(b-d) unless we're dividing by zero, the above are also equal to:

$\frac{y-z}{xz-xy}$ = $\frac{y-z}{x(z-y)}$ = $\frac{-1}{x}$

which is negative, whereas the first three fractions must be positive. So it's a contradiction. Therefore zx-xy must have been zero, so z=y.

Similarly, x=y

x = y = z therefore x-y = 0 or y-x =0 and so on............
by solving above eq. we get ....
(x-y)(x+y) = y(z-x)........
Sol. 2 x =y = z therefore x/y = 1
given eq...... x(x+y) = y(y+ z)
(x/y)(x+y) = (y+z)
x = z {x/y =1]