Find all the values of the parameter 'a' for which the given inequality is satisfied for all real values of x. Find all the values of the parameter 'a' for which the inequality is satisfied for all real values of x.
$$a\cdot 9^x+4\cdot \left(a-1\right)\cdot 3^x+\left(a-1\right)>0$$
My attempt is as follows:-
$$a\cdot \left(9^x+4\cdot 3^x+1\right)-(4\cdot 3^x+1)>0$$
$$a\cdot \left(9^x+4\cdot 3^x+1\right)>4\cdot 3^x+1$$
$$a>\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$$
Now if a is greater than the maximum value of $\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$, then inequality will be true for all x.
So if we can find the range of $\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$, then we can say a should be greater than the maximum value in the range.
Let's assume y=$\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$ and substitute $3^x$ with $t$.
$$y=\frac{4t+1}{t^2+4t+1}$$
$$yt^2+4ty+y=4t+1$$
$$yt^2+4t(y-1)+y-1=0$$
We want to have real values of t satisfying the equation, so $D>=0$
$$16(y-1)^2-4*y*\left(y-1\right)>=0$$
$$4(y-1)(4y-4-y)>=0$$
$$4(y-1)(3y-4)>=0$$
$$y\in \left(-\infty,1 \right] \cup \left[\frac{4}{3},\infty\right)$$
So I am getting maximum value tending to $\infty$ for y=$\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$
I am not able to understand where am I making mistake.
Official answer is $a\in \left[1,\infty\right) $

I think I got the point where I did mistake, thanks to @copper.hat
It is important to note that when substituting $3^x$ with $t$, it means t will also be greater than $0$.
So we have to find the condition where at least one root is positive, so there can be multiple cases to it.
Case 1: When both the roots are positive, it means product of roots should be positive and sum of roots should be positive, so from the equation $yt^2+4t(y-1)+y-1=0$ , we can say $\frac{y-1}{y}>0$ and $\frac{-4\left(y-1\right)}{y}>0$, so $y\in \phi$.
Case 2: When one root is positive and other root is negative, it means product of roots should be negative, so from the equation $yt^2+4t(y-1)+y-1=0$, we can say $\frac{y-1}{y}<0$, so $y\in (0,1)$
Case 3: When one root is zero and other root is positive, it means product of roots should be zero and sum of roots should be positive, $\frac{y-1}{y}=0$ and $\frac{-4\left(y-1\right)}{y}>0$, so $y=1$ and $y\in \left(0,1\right)$, so $y\in \phi$
Taking union of all conditions in case $1,2,3$ , we get $y\in (0,1)$, so maximum value is tending to $1$ but not $1$. Hence $a \in \left[1,\infty\right)$
