<s>[INST] 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.
 [/INST] We can obtain $yz+zx+xy=2$ simply by finding the values of $x$, $y$, and $z$. We are given $$y^2+yz+z^2=1,\qquad(1)$$$$z^2+zx+x^2=4,\qquad(2)$$$$x^2+xy+y^2=3,\,\qquad(3)$$with $x,y,z>0$. Subtracting eqn $3$ from eqn $2$, and noticing the factor $z-y$ on the LHS, gives $$(z-y)(x+y+z)=1.$$Similarly, from eqn $2$ minus eqn $1$, we get$$(x-y)(x+y+z)=3.$$The ratio of the latter two equations yields $x-y=3(z-y)$, or$$x=3z-2y.$$ Now substituting for $x$ from the above equation into eqn $3$ yields$$y^2-3yz+3z^2=1.$$By subtracting this from eqn $1$, we find $2yz-z^2=0$, or$$z=2y$$(since $z\neq0$). Substitution for $z$ into eqn $1$ now gives $7y^2=1$. It follows that $y=1/\surd7$, because $y>0$,  and then $z=2/\surd7$ and $x=4/\surd7$, giving$$yz+zx+xy=2.$$
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