Determining the range of $6^n+ 6^{-n} +3^n +3^{-n}+2$ 
I have to solve for range of the function $$6^n+ 6^{-n} +3^n +3^{-n}+2$$ 

The textbook solves it as 
$$\left(\sqrt{6^n} -\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} -\sqrt{ 3^{-n}} \right)^2 +6 \tag{1}$$ 
i.e., $$(a-b)^2+(a-b)^2$$ 
which will always be greater than $6$, so the range is $(6,\infty)$ (since other two terms are squared). But, if we take 
$$\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 +2 \tag{2}$$ 
or $$\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 +2 \tag{3}$$
instead of $(1)$, we get that the range is $(-2,\infty)$ or $( 2,\infty)$, respectively.

So, how do we know what range is correct?

 A: If you'll take $$3^n+3^{-n}+6^n+6^{-n}+2=\left(\sqrt3^n+\sqrt3^{-n}\right)^2+\left(\sqrt6^n+\sqrt6^{-n}\right)^2-2\geq-2,$$ but the equality does not occur.
But in the following writing 
 $$3^n+3^{-n}+6^n+6^{-n}+2=\left(\sqrt3^n-\sqrt3^{-n}\right)^2+\left(\sqrt6^n-\sqrt6^{-n}\right)^2+6\geq6$$ the equality occurs for $n=0,$ which says that $6$ is a minimal value. 
A: Note that:
$$6^n+ 6^{-n} +3^n +3^{-n}+2=\left(\sqrt{6^n} -\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} -\sqrt{ 3^{-n}} \right)^2 +6 \ \ \text{(textbook)}\\
6^n+ 6^{-n} +3^n +3^{-n}+2=\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 \overbrace{\require{cancel}\cancel{\color{red}+}}^{-} 2\ \ \text{(yours)}\\
$$
Note that by AM-GM:
$$x+\frac1x\ge 2, x>0,$$
the equality occurs for $x=1$.
Hence:
$$6^n+6^{-n}\ge 2, 3^n+3^{-n}\ge 2,\\
\sqrt{6^n}+\sqrt{6^{-n}}\ge 2,\sqrt{3^n}+\sqrt{3^{-n}}\ge 2,$$
the equality in each of the four inequalities occurs for $n=0$.
Thus, your method must be:
$$\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 -2 \ge 6,$$
the equality occurs for $n=0$.
