Parametric exponential inequality Find the values of $m$ s.t.
$$
\left(\frac{9}{25}\right)^x-m\left(\frac{3}{5} \right)^x+1>0,
$$ for all $x<0$.
My attempt is the following: let $y=(3/5)^x>1$ and the inequality transforms as follows
$$
h(y)=y^2-my+1>0, \quad \text{ for all } y>1.
$$
Now I made an analysis of second degree polynomial function in variable $y$. So $h(y)<0,\text{for all } y>1$, if the discriminant of $h$ is strictly negative. Hence $$
m^2-4<0.
$$ So, $m$ must be a number from $(-2,2)$. I don't know how to use the information that $x<0$ and if my solution is complete. 
Thank you!
 A: Your work up to
$$h(y) = y^2 - my + 1 \gt 0, \quad \text{ for all } y \gt 1 \tag{1}\label{eq1}$$
is completely correct. However, your next statement that $h(y) \lt 0$ for all $y \gt 1$ applies if the discriminant is strictly negative is not completely true. For example, as Gibbs states in the comment, if $m \lt 0$ (including if $m \lt -2$ so the discriminant is negative), then \eqref{eq1} is always positive for $y \gt 1$. There are several methods to prove \eqref{eq1}, such as those shown below.

I believe the easiest way to determine the values of $m$ which work is to use
$$m = 2 - z \tag{2}\label{eq2}$$
in \eqref{eq1} to get
$$y^2 - (2 - z)y + 1 = y^2 - 2y + 1 + zy = (y - 1)^2 + zy \gt 0 \tag{3}\label{eq3}$$
To investigate $z \lt 0$, use $y = 1 - \frac{z}{2} \gt 1$ in \eqref{eq3} to get
$$\left(-\frac{z}{2}\right)^2 + z\left(1 - \frac{z}{2}\right) = \frac{z^2}{4} + z - \frac{z^2}{2} = -\frac{z^2}{4} + z \lt 0 \tag{4}\label{eq4}$$
This shows $z \lt 0$ doesn't work. For $z \ge 0$, with $y \gt 1$, then $(y - 1)^2 \gt 0$ and $zy \ge 0$, so \eqref{eq3} always holds. This shows, using \eqref{eq2}, that any $m \le 2$ will work for \eqref{eq1}.

An alternate way to check is by using the discriminant of \eqref{eq1} where the left side is equal to $0$, i.e.,
$$d = m^2 - 4 \tag{5}\label{eq5}$$
where, if $d \ge 0$, then the zeros are given by
$$y = \frac{m \pm \sqrt{m^2 - 4}}{2} \tag{6}\label{eq6}$$
Note that for \eqref{eq1} to always hold, since $\lim_{y \to \infty} h(y) = \infty$, then the left side of \eqref{eq1} must be $\ge 0$ for $y = 1$ and either it doesn't have any zeroes, i.e., $d \lt 0$, or $d \ge 0$ but the zero(s) in \eqref{eq6} occur only where $y \le 1$. For the first part, with $y = 1$, the left side of \eqref{eq1} becomes
$$1 - m + 1 = 2 - m \ge 0 \; \; \to \; \; m \le 2 \tag{7}\label{eq7}$$
For $m = 2$, then $y = 1$ is the only root, with the left side of \eqref{eq1} becoming $(y-1)^2$ so it's always $\gt 0$ for $y \gt 1$. For $-2 \lt m \lt 2$, \eqref{eq5} gives $d \lt 0$, so \eqref{eq1} has no roots. For $m \le -2$, it's clear that \eqref{eq1} is always true due the left side adding $3$ positive values. However, you can also note, for $m \le -2$, that $\sqrt{m^2 - 4} \lt \sqrt{m^2} = \left|m\right|$ so $m \pm \sqrt{m^2 - 4} \lt 0$. Thus, in \eqref{eq6}, $y \lt 0$ so the zeros occur outside of the range $y \gt 1$ and, thus, \eqref{eq1} will always hold.

In summary, any of the above methods can be used to show that, for all $x \lt 0$, your original equation of
$$\left(\frac{9}{25}\right)^x - m\left(\frac{3}{5} \right)^x + 1 \gt 0 \tag{8}\label{eq8}$$
is true for any $m \le 2$.
