<s>[INST] For which $a$, equation $4^x-a2^x-a+3=0$ has at least one solution. 
Find all values of $a$ for which the equation $4^x-a2^x-a+3=0$ has at least one solution.

$\bf{My\; Try::}$ We can write it as $$2^{x}-a-\frac{a}{2^x}+\frac{3}{2^x}=0$$
So $\displaystyle \left(2^x+\frac{3}{2^x}\right)=a\left(1+\frac{1}{2^x}\right).$
Now for the existance of solution $\displaystyle 2^x+\frac{3}{2^x}\geq 2\sqrt{3}$ Using $\bf{A.M\geq G.M}$
So $\displaystyle a\left(1+\frac{1}{2^x}\right)\geq 2\sqrt{3}\Rightarrow a\geq \sqrt{3}\cdot \frac{2^x}{2^x+1}\geq \sqrt{3}$
But answer given as $a\geq 2,$ please explain me whats wrong with that, Thanks
 [/INST] You found a necessary condition, but not a sufficient one. You've concluded that
$$a \ge \frac{2^x + \frac{3}{2^x}}{1 + \frac 1 {2^x}}$$
Your application of AM-GM is all about minimizing the numerator, but it doesn't minimize the entire fraction. In fact, AM-GM is sharp when $2^x = \sqrt{3}$, in which case the fraction is $6 / (\sqrt{3} + 1) \approx 2.2$. 

To minimize the whole thing: Recognize that it's equivalent to 
$$a \ge \frac{4^x + 3}{2^x + 1}$$
As $x \to \infty$, this blows up; as $x \to -\infty$, this tends to $3$. Standard calculus techniques (e.g. root of first derivative) shows that this is minimized when $x = 0$, and the minimum is in fact $2$. By continuity, each value of $a \ge 2$ has a corresponding $x$. Hence $a \ge 2$ is sufficient.
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