<s>[INST] 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) $
 [/INST] 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 $a\geq1.$
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