Real positive solution to specific system of equations

Question

In the course of an optimization problem, I have encountered the following system of equations for $\alpha, \beta > 0$: $$ \begin{align} a_1 + a_2 + b_1 + b_2 &= \alpha + \beta\\ a_1b_1 + \frac{1}{4} a_2 b_2 &= \alpha \beta. \end{align} $$ Is there a general approach to tackle such a system and does this system have a real solution with $a_1, a_2 , b_1, b_2 > 0$?

Answer 1

Note that $x + y = s$, $x y = t$ has solutions $$ x = \frac{s - \sqrt{s^2 - 4 t}}{2},\ y = \frac{s + \sqrt{s^2 - 4 t}}{2} $$ which are positive if and only if $s^2 \ge 4 t$, and $s,t>0$. Of course $(\alpha +\beta)^2 \ge 4 \alpha \beta$, with equality iff $\alpha = \beta$.

If $\alpha \ne \beta$, we can choose any $a_2$ and $b_2$ that are positive but small enough that $s > 0$, $t> 0$ and $s^2 > 4 t$ where $s = \alpha + \beta - a_2 - b_2$ and $t = \alpha \beta - a_2 b_2/4$. Then we have positive solutions $a_1 = (s - \sqrt{s^2 - 4 t})/2$, $b_1 = (s + \sqrt{s^2 - 4 t})/2$.

If $\alpha = \beta$, the system is not solvable with positive $a_1,a_2,b_1, b_2$, because squaring the first equation and subtracting $4$ times the second gives $$ \left( a_{{1}}-b_{{1}} \right) ^{2}+2\,a_{{1}}a_{{2}}+2\,a_{{1}}b_{{2 }}+{a_{{2}}}^{2}+2\,b_{{1}}a_{{2}}+a_{{2}}b_{{2}}+2\,b_{{1}}b_{{2}}+{b _{{2}}}^{2} =0$$