Question

Find the values of the parameter $a$ for which the inequality ${a9}^x+4(a-1)3^x+a>1$ is satisfied for all real values of $x$

Answer 1

Given inequality

$a{9}^x+4(a-1)3^x+a>1$

Put $t=3^x$, in the original equation we get
${at}^2+4(a-1)t+a>1$

$\Rightarrow {at}^2+4(a-1)t+(a-1)>0(t>0,\because 3^x>0)$

This is possible in two cases.

Case I: The parabola $f\left(t\right)={at}^2+4(a-1)t+(a-1)$ opens upward, with its vertex lying in the non positive part of the t-axis, as shown in the following four figures.
$\therefore a>0$ ; sum of roots $\le 0$ and $f\left(0\right)\ge 0$
$\Rightarrow -\frac{4(a-1)}{2a}\le 0$ and $a-1\ge 0$

$\therefore a>0,a-1\ge 0$ and $a\ge 1$

Hence $a\ge 1$ .........(1)

Case II: The parabola $f\left(t\right)$ opens upward, with its vertex lying in the positive direction of t-axis then

$a>0$ ; sum of roots >0 and $D\le 0$
$a>0,-\frac{4(a-1)}{2a}>0$ and $16{(a-1)}^2-4(a-1)\le 0$

$\therefore a>0,a<1$ and $4(a-1)(3a-4)\le 0$
From wavy curve method, these inequalities cannot occur simultaneously.

No value of $a$ satisfies all these.

Hence $a\ge 1$ from (1).