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) $
 A: Prove that $$\frac{4\cdot3^x+1}{9^x+4\cdot3^x+1}<1$$ for any real $x$.
Now, $$\lim_{x\rightarrow-\infty}\frac{4\cdot3^x+1}{9^x+4\cdot3^x+1}=1,$$ which gives $$\sup\frac{4\cdot3^x+1}{9^x+4\cdot3^x+1}=1$$ and the answer $a\geq1.$
A: There is an easier way :-):
Let $t = 3^x$, noting that we need $t>0$. The equation becomes
$at^2+4(a-1)t +(a-1) > 0$ for all $t>0$.
First note that if the above is true for all $t >0$ then we must have (taking
the limit as $t \downarrow 0$) that $(a-1) \ge 0$ (note $\ge$ not $>$).
In
particular, $a \ge 1$ must hold.
The $\min$ of the left hand side (over all $t$) can be found using:
$2at + 4(a-1) = 0$, or $t^* = -2 {a-1\over a}$.
Since $t^* \le 0$, we see (because it is a quadratic) that the 
left hand side is an increasing function of $t$ for $t \ge 0$ and the 
value at $t=0$ is $a-1$.
In particular, if $t >0$ we have $at^2+4(a-1)t +(a-1) > a-1\ge 0$.
A: 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 $a \in \left[1,\infty\right)$
