<s>[INST] Solve system:$yz(y+z-x)= a(x+y+z)\\ zx(z+x-y)= b(x+y+z)\\ xy(x+y-z)= c(x+y+z)$ Solve system:
$$yz(y+z-x)= a(x+y+z)\\
 zx(z+x-y)= b(x+y+z)\\
 xy(x+y-z)= c(x+y+z)$$
$$a, b, c>0$$
I can only solve the system at $a= 2, b= 5, c= 10$. I have tried all things but it's hard with me. Somebody help me?
 [/INST] From the first equation
$$y z \left(( x + y + z ) - 2 x \right) = a ( x + y + z ) \quad\to\quad
-2x y z = ( a - y z )( x + y + z ) \tag{$\star$}$$
Likewise,
$$-2xyz=(b-zx)(x+y+z) \qquad -2 x y z = ( c - x y )( x + y + z )$$
If, say, $x=0$, then $(a-yz)(y+z) = b(y+z) = c(y+z)=0$; thus, either $y=-z$, or $b=c=0$ and $a=yz$ (or both). Similarly for $y=0$ and $z=0$.
Going forward, we take $x$, $y$, $z$ non-zero, noting that $x+y+z$ is then also non-zero. Consequently, we can write
$$a - y z = b - z x = c - x y = k$$
for some (non-zero) $k$. Therefore,
$$x^2 = \frac{xy\cdot zx}{yz}= \frac{(b-k)(c-k)}{a-k} \qquad y^2 = \frac{(c-k)(a-k)}{b-k} \qquad z^2=\frac{(a-k)(b-k)}{c-k}$$
Since $x^2$, $y^2$, $z^2$ are positive (we've handled the zero case), we see that an even number of $a-k$, $b-k$, $c-k$ can be negative. Let's take $a-k$ to be positive, and define $s = \pm1$ to have the common sign of $b-k$ and $c-k$. Then,
$$ x = \frac{r}{a-k} \qquad y = \frac{sr}{b-k} \qquad z = \frac{sr}{c-k} \qquad r := \sqrt{(a-k)(b-k)(c-k)} \tag{$\star\star$}$$
Since
$$b = k + z x = k + \frac{s(a-k)(b-k)(c-k)}{(c-k)(a-k)} = k + s(b-k)$$
we have that $s$ is unambiguously $1$. Substituting $x$, $y$, $z$ into $(\star)$, and reducing, ultimately gives

$$k^3 - k( b c + c a + a b ) + 2 a b c = 0 \tag{$\star\star\star$}$$

While there is a Cubic Formula for solving such equations explicitly, we won't bother. (Actually, see below.)
However, we might as well solve the specific case of $a=2$, $b=5$, $c=10$ as a sanity check. We can factor $(\star\star\star)$ to get
$$(k+10)(k^2-10k+20) = 0 \quad\to\quad k = -10\;\text{or}\; 5 \pm \sqrt{5}$$
Since all of $a-k$, $b-k$, $c-k$ must be positive, $a=2$ eliminates $k=5\pm\sqrt{5}$ from consideration; therefore,

$$x = \frac{\sqrt{12\cdot 15\cdot 20}}{12} = \frac{60}{12} = 5 \qquad y = 4 \qquad z = 3$$


Edit. The question has been updated to ask for proof that a positive solution exists for positive $a$, $b$, $c$.
First of all, the parameters being strictly positive allow us to say specifically that, if one of $x$, $y$, $z$ are zero, then the other two are mutual negatives. So, the trivial solutions are slightly streamlined.
For non-zero $x$, $y$, $z$, the previous argument holds as-is. Considering $(\star\star\star)$ with $k\to-k$, we can invoke the Descartes Rule of Signs, counting just one sign change in the coefficient sequence $\{-1, bc + ca + ab, 2abc\}$. Thus, the polynomial has one strictly-negative root $k$, so that $a-k$, $b-k$, $c-k$ are positive, as are the corresponding $x$, $y$, $z$. $\square$   

In the interest of completeness, I'll give the explicit roots of $(\star\star\star)$. Since the equation involves a "depressed" cubic (that is, its second-degree term is zero), we can jump to the trigonometric form.
$$k = 2 \sqrt{\frac13(a b + b c + c a )} 
 \;\cos\left(\frac{\pi}{3} ( 2 n-1 ) + \frac13 \operatorname{arccos}\frac{3^{3/2} a b c}{( a b + b c + c a )^{3/2}} \right)
$$
with $n = 0, 1, 2$. The Arithmetic-Geometric Mean Inequality guarantees that the argument of the inverse cosine doesn't exceed $1$, so that all roots are real. The case $n=2$ reduces to 
$$k = -2 \sqrt{\frac13(a b + b c + c a )} 
 \;\cos\left(\frac13 \operatorname{arccos}\frac{3^{3/2} a b c}{( a b + b c + c a )^{3/2}}\right)
$$
giving the negative $k$ that guarantees $a-k$, $b-k$, $c-k$, and $x$, $y$, $z$, are positive.
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