User talk:Jhoshen1/Sandbox


 * $\omega _\alpha ^2 = bc - \dfrac { {b c a^2 }} {\paren { b + c}^2 }$


 * $\omega _\beta ^2 = ac - \dfrac { {a c b^2 }} {\paren {a + c}^2 }$

Equating the two equations, we have:


 * $ bc - \dfrac { {b c a^2 }} {\paren { b + c}^2 } = ac - \dfrac { {a c b^2 }} {\paren {a + c}^2 } $


 * $\leadsto c(b - a) = \dfrac { {b c a^2 }} {\paren { b + c}^2 } - \dfrac { {a c b^2 }} {\paren {a + c}^2 } $

Multiplying throughout by $(b + c)^2 (a + c)^2$
 * $\leadsto c(b - a)(b + c)^2 (a + c)^2 = bca^2 (a + c)^2  - acb^2 (b + c)^2 $
 * $\leadsto c(b - a)(b + c)^2 (a + c)^2 = ca \paren { a(a + c)^2  - b(b + c)^2 } \;\;\; (1) $

Substituting $b$ for $a$ in $(1)$ proves that $b=a$ is a solution for $(1)$ (both sizes of the equation evaluate to zero).

We still have to show that $b=a$ is the only solution for $(1)$. We do that by assuming that $b>a$. Because $ c(b - a)(b + c)^2 (a + c)^2 > 0$, the left hand side of $(1)$ is positive. Because $b > a $ and $b+c > a+c $
 * $\leadsto a(a + c)^2 < b(b + c)^2$,

the right hand side of $(1)$ is negative, which is a contradiction. Similarly, we also get a contradiction if we assume $b< a$. This assumption yields a negative left hand side and a positive right hand side for $(1)$. To sum up, $b=a$ is the only solution for $(1)$