Find $xy+yz+zx$ given systems of three homogenous quadratic equations for $x, y, z$ This is a question from Math Olympiad.

If $\{x,y,z\}\subset\Bbb{R}^+$ and if $$x^2 + xy + y^2 = 3 \\ y^2 + yz + z^2 = 1 \\ x^2 + xz + z^2 = 4$$ find the value of $xy+yz+zx$.

I basically do not know how to approach this question. Please let me know how to approach this question, and if you attach full explanation, I will appreciate it. Thanks.

Suppose $x,y,z > 0$ satisfy the system
$$
\begin{cases}
x^2 + xy + y^2 = 3&\;\;\;(\text{eq}1)\\[4pt]
y^2 + yz + z^2 = 1&\;\;\;(\text{eq}2)\\[4pt]
z^2 + zx + x^2 = 4&\;\;\;(\text{eq}3)\\
\end{cases}
$$
From $(\text{eq}2)$, since $y,z > 0$, we get $y,z < 1$, hence from $(\text{eq}1)$, we must have $x > 1$. 

Let $a = x + y + z$, and let $b = xy + yz + zx$.

The goal is to find the value of $b$.

Since $x,y,z > 0$, and $x > 1$, we get $a > 1$.

Identically, we have $x^2 + y^2 + z^2 = a^2 -2b$.

From the sum $(\text{eq}1)+ (\text{eq}2) +(\text{eq}3)$, we get
$$2(x^2 + y^2 + z^2) + (xy + yz + zx) = 8$$
hence 
$$2a^2-3b=8\qquad(\text{eq}4)$$

Subtracting $(\text{eq}2)$ from $(\text{eq}1)$, we get
\begin{align*}
&(x^2 +xy + y^2) - (y^2 + yz + z^2) = 2\\[4pt]
\implies\;&(x^2 - z^2) + (xy - yz) = 2\\[4pt]
\implies\;&(x-z)(x+y+z) = 2\\[4pt]
\implies\;&x-z =\frac{2}{a}\\[4pt]
\implies\;&z^2-2zx+x^2 = \frac{4}{a^2}\qquad(\text{eq}5)\\[4pt]
\end{align*}
Subtracting $(\text{eq}2)$ from $(\text{eq}3)$, we get
\begin{align*}
&(z^2 +zx + x^2) - (y^2 + yz + z^2) = 3\\[4pt]
\implies\;&(x^2 - y^2) + (zx - yz) = 3\\[4pt]
\implies\;&(x-y)(x+y+z) = 3\\[4pt]
\implies\;&x-y =\frac{3}{a}\\[4pt]
\implies\;&x^2-2xy+y^2 = \frac{9}{a^2}\qquad(\text{eq}6)\\[4pt]
\end{align*}
Subtracting $(\text{eq}1)$ from $(\text{eq}3)$, we get
\begin{align*}
&(z^2 +zx + x^2) - (x^2 + xy + y^2) = 1\\[4pt]
\implies\;&(z^2 - y^2) + (zx - xy) = 1\\[4pt]
\implies\;&(z-y)(x+y+z) = 1\\[4pt]
\implies\;&z-y =\frac{1}{a}\\[4pt]
\implies\;&y^2-2yz+z^2 = \frac{1}{a^2}\qquad(\text{eq}7)\\[4pt]
\end{align*}

From the sum $(\text{eq}5)+ (\text{eq}6) +(\text{eq}7)$, we get
$$2(x^2 + y^2 + z^2) - 2(xy + yz + zx) = \frac{14}{a^2}$$
hence 
$$a^2-3b=\frac{7}{a^2}\qquad(\text{eq}8)$$
Subtracting $(\text{eq}8)$ from $(\text{eq}4)$, we get
\begin{align*}
&(2a^2-3b) - (a^2-3b) = 8-\frac{7}{a^2}\\[4pt]
\implies\;&a^2 = 8-\frac{7}{a^2}\\[4pt]
\implies\;&a^4 - 8a^2 + 7=0\\[4pt]
\implies\;&(a^2-1)(a^2-7)=0\\[4pt]
\implies\;&a^2 = 7\qquad\text{[since $a > 1$]}\\[4pt]
\end{align*}
Then from $(\text{eq}4)$, we have
$$b = \frac{2a^2-8}{3}$$
hence
$$b = \frac{2(7)-8}{3} = 2$$
