I have a quadratic equation with unknown $x$ given as (all parameters are positive): $$ a_{1}(x+m_{1}\frac{x^{2}}{2})-a_{2}(\tau+m_{2}\frac{\tau^{2}}{2})=0 $$ I want to show that the positive solution satisfies $$ x\geq\frac{a_{2}m_{2}}{a_{1}m_{1}}\tau $$ assuming that $$ a_{1}(\tau+m_{1}\frac{\tau^{2}}{2})-a_{2}(\tau+m_{2}\frac{\tau^{2}}{2})\geq0 $$ and $a_{1}m_{1}\geq a_{2}m_{2}$.
What I've tried: If we write the solution to the quadratic equation we get: $$ x=\frac{-a_{1}+\sqrt{a_{1}^{2}+2a_{1}m_{1}(a_{2}\tau+a_{2}m_{2}\frac{\tau^{2}}{2})}}{a_{1}m_{1}} $$ The term in the square root can be written as \begin{eqnarray*} a_{1}^{2}+2a_{1}m_{1}(a_{2}\tau+a_{2}m_{2}\frac{\tau^{2}}{2}) & = & (a_{1}+a_{2}m_{2}\tau)^{2}-(a_{2}m_{2}\tau)^{2}-2a_{1}a_{2}m_{2}\tau\\ & & +2a_{1}m_{1}a_{2}\tau+a_{1}m_{1}a_{2}m_{2}\tau^{2} \end{eqnarray*} so if we show that $$ -(a_{2}m_{2}\tau)^{2}-2a_{1}a_{2}m_{2}\tau+2a_{1}m_{1}a_{2}\tau+a_{1}m_{1}a_{2}m_{2}\tau^{2}\geq0 $$ we are done. But I haven't been able to reduce the above to the assumption. What I get is instead \begin{eqnarray*} -(a_{2}m_{2}\tau)^{2}-2a_{1}a_{2}m_{2}\tau+2a_{1}m_{1}a_{2}\tau+a_{1}m_{1}a_{2}m_{2}\tau^{2} & = & \tau^{2}a_{2}m_{2}(a_{1}m_{1}-a_{2}m_{2})+2\tau(a_{1}m_{1}a_{2}-a_{1}m_{2}a_{2})\\ & \geq & \tau^{2}a_{2}m_{2}(a_{1}m_{1}-a_{2}m_{2})+2\tau a_{2}m_{2}(a_{2}-a_{1}) \end{eqnarray*} which is not necessarily positive, but I believe that the inequality must be still valid.
We note that from solving your equation in $\tau$, $$\tau \geq \frac{2(a_2-a_1)}{a_1m_1-a_2m_2}$$ Then, we are looking to show that $$-a_1+\sqrt{a_1^2+2a_1m_1\left(a_2\tau+a_2m_2\frac{\tau^2}{2}\right)} \geq a_2m_2\tau$$ In the case that $a_2 > a_1$ (and $m_1 > m_2$), we note that $$a_2\tau+a_2m_2\frac{\tau^2}{2} \geq \frac{2a_2(a_2-a_1)}{a_1m_1-a_2m_2}+\frac{2a_2(a_1-a_2)^2m_2}{(a_1m_1-a_2m_2)^2} = \frac{2a_1a_2(a_2-a_1)(m_1-m_2)}{(a_1m_1-a_2m_2)^2}$$ Therefore, \begin{align*} a_1^2+2a_1m_1\left(a_2\tau+a_2m_2\frac{\tau^2}{2}\right) &\geq a_1^2+\frac{4a_1^2a_2(a_2-a_1)m_1(m_1-m_2)}{(a_1m_1-a_2m_2)^2} \\ &= \frac{a_1^2(a_1m_1-2a_2m_1+a_2m_2)^2}{(a_1m_1-a_2m_2)^2} \end{align*} We can see that as $a_1m_1 \geq a_2m_2$, $$a_1m_1-2a_2m_1+a_2m_2 \leq 2a_1m_1-2a_2m_1 = 2m_1(a_1-a_2) < 0$$ so $$\sqrt{\frac{a_1^2(a_1m_1-2a_2m_1+a_2m_2)^2}{(a_1m_1-a_2m_2)^2}} = -\frac{a_1(a_1m_1-2a_2m_1+a_2m_2)}{a_1m_1-a_2m_2}$$ so finally, \begin{align*} -a_1+\sqrt{a_1^2+2a_1m_1\left(a_2\tau+a_2m_2\frac{\tau^2}{2}\right)} &\geq -a_1-\frac{a_1(a_1m_1-2a_2m_1+a_2m_2)}{a_1m_1-a_2m_2} \\ &= \frac{2a_1(a_2-a_1)m_1}{a_1m_1-a_2m_2} \\ &= a_1m_1\tau\end{align*} As $a_1m_1 \geq a_2m_2$, we therefore have that $a_1m_1\tau \geq a_2m_2\tau$, so we are done with the first case.
Note that the inequality is not necessarily true in the case that $a_1 > a_2$. A counterexample, found by Mathematica, could be $a_1 = 8$, $a_2 = 1$, $m_1 = 1$, $m_2 = 5$, and $\tau = 2$. This gives us the positive solution $x = 1$, but $\frac{a_2m_2}{a_1m_1}\tau = \frac{5}{4}$.