Let us consider the following system of two polynomial equations of third order for two real numbers $x_1,x_2$: $$x_i (x_i + 2) (x_i + 4) - 2 a_i (x_1 + x_2 + 4) = 0,$$ $i =1,2$. Here $a_1 >0$ and $a_2 >0$. It is necessary to prove that for any set of positive numbers $(a_1,a_2)$ there exists a unique solution of the system obeying $x_1 > 0$ and $x_2 >0$. In this case one can readily prove that the functions $x_i = x_i(a_1,a_2)$ ($i =1,2$) are smooth in ${\mathbb R}_{+}^2$. Remark: for $x_1 + x_2 + 4 \neq 0$ the summing of two equations leads us to the relation $x_1^2 + x_2^2 - x_1 x_2 + 2 (x_1 + x_2) = 2 (a_1 + a_2)$.


Solving the first equation for $x_2$ yields $$ x_{{2}}={\frac {{x_{{1}}}^{3}}{2\;a_{{1}}}}+3\,{\frac {{x_{{1}}}^{2}}{a_{{1} }}}-{\frac { \left( a_{{1}}-4 \right) x_{{1}}}{a_{{1}}}}-4 $$ This cubic is $0$ at $x_1 = -4$ and $-1 \pm \sqrt{1+2 a_1}$. It is convex and increasing for $x_1 > -1 + \sqrt{1+2 a_1}$. Similarly, for the second equation, with indices $1$ and $2$ interchanged. From this it is easy to see that the two curves intersect exactly once in the first quadrant.

  • $\begingroup$ But solving the first equation for $x_2$ should give us a cubic polynomial in $x_1$. $\endgroup$ – Vladimir Feb 2 '17 at 16:06
  • $\begingroup$ Sorry, fixed it. I don't know how the wrong formula crept in, but the conclusion is the same. $\endgroup$ – Robert Israel Feb 2 '17 at 17:45
  • $\begingroup$ Dear Robert Israel, thank you very much! Your proof is done in a geometric style which is close (to my opinion) to that of V.I. Arnol'd. $\endgroup$ – Vladimir Feb 2 '17 at 19:50

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