What is the minimum standard deviation for a normal PDF such that one tail is always larger than that of a second normal PDF (different means)?

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Say I have two weighted normal distributions, $$ f_1(x) = \frac{a}{2 \sigma_1} e^{-\frac{(x-\mu_1)^2}{2\sigma_1^2}} $$ and $$ f_2(x) = \frac{1-a}{2 \sigma_2} e^{-\frac{(x-\mu_2)^2}{2\sigma_2^2}} $$ Let's also say that $\mu_2>\mu1$. For fixed values of $\mu_1$, $\sigma_1$, and $\mu_2$, and fixed relative weights $a$ and $1-a$, what is the minimum value of $\sigma_2$ such that $f_2 > f_1$ for all $x>\mu_2$? In other words, if I want to make a normal PDF with a right tail that is always greater than the right tail of another normal PDF, what are my constraints on the standard deviation?

I'd settle for some approximations too, if that helps.