Determining the range of $6^n+ 6^{-n} +3^n +3^{-n}+2$

Question

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?

Answer 1

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.

Answer 2

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 $n=0$.

Answer 3

Because

\begin{align*} &\color{red}{\left(\sqrt{6^n} -\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} -\sqrt{ 3^{-n}} \right)^2 +6} \\&=(\sqrt{6^n})^2-2\sqrt{6^n}\sqrt{ 6^{-n}}+(\sqrt{ 6^{-n}})^2+(\sqrt{3^n})^2-2\sqrt{3^n}\sqrt{ 3^{-n}}+(\sqrt{ 3^{-n}})^2 +6\\ &= 6^n-2\sqrt{6^n 6^{-n}}+6^{-n}+3^n-2\sqrt{3^n 3^{-n}}+ 3^{-n}+6\\ &= 6^n-2\sqrt{6^{n-n}}+6^{-n}+3^n-2\sqrt{3^{n-n}}+ 3^{-n}+6\\ &= 6^n-2\sqrt{6^{0}}+6^{-n}+3^n-2\sqrt{3^{0}}+ 3^{-n}+6\\ &= 6^n-2\sqrt{1}+6^{-n}+3^n-2\sqrt{1}+ 3^{-n}+6\\ &= 6^n-2+6^{-n}+3^n-2+ 3^{-n}+6\\ &= 6^n+6^{-n}+3^n+ 3^{-n}+6-2-2\\ &= 6^n+6^{-n}+3^n+ 3^{-n}+6-4\\ &\color{magenta} =\color{red}{6^n+ 6^{-n} +3^n +3^{-n}+2} \end{align*}

and

\begin{align*} \color{blue}{\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 +2} &=6^n+2+6^{-n}+3^n+2+3^{-n}+2 \\&=6^n+6^{-n}+3^n+3^{-n}+6 \\&\color{magenta} \neq \color{red}{6^n+ 6^{-n} +3^n +3^{-n}+2} \end{align*}

This means:

If $$\color{red}{\left(\sqrt{6^n} -\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} -\sqrt{ 3^{-n}} \right)^2 +6} = \color{red}{6^n+ 6^{-n} +3^n +3^{-n}+2}$$ and $$\color{blue}{\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 +2} \neq \color{red}{6^n+ 6^{-n} +3^n +3^{-n}+2},$$ then $$ \color{blue}{\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 +2} \neq \color{red}{\left(\sqrt{6^n} -\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} -\sqrt{ 3^{-n}} \right)^2 +6} $$

But

$$ \color{red}{\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 -2} = \color{red}{6^n+ 6^{-n} +3^n +3^{-n}+2} $$

and

$$ \color{red}{\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 -2} = \color{red}{\left(\sqrt{6^n} -\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} -\sqrt{ 3^{-n}} \right)^2 +6}$$

The range of (or the values given by the following expression for $n=0,1,2,3,\cdots$) $$\color{red}{6^n+ 6^{-n} +3^n +3^{-n}+2}$$ can be seen better, when you look at $$ \color{red}{\left(\sqrt{6^n} -\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} -\sqrt{ 3^{-n}} \right)^2 +6}$$ instead of looking at $$\color{red}{\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 -2}$$

Looking at $ \color{red}{\left(\sqrt{6^n} -\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} -\sqrt{ 3^{-n}} \right)^2 +6}$ for $n=0$ gives immediately a value of $6$. For values $n \geq 1$, we see that $\sqrt{6^n} > \sqrt{ 6^{-n}}$ and $\sqrt{3^n} > \sqrt{ 3^{-n}}$. So we have for $n \geq 1$ that $\sqrt{6^n} - \sqrt{ 6^{-n}}>0$ and $\sqrt{3^n} -\sqrt{ 3^{-n}}> 0$. Also $\left(\sqrt{6^n} -\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} -\sqrt{ 3^{-n}} \right)^2 >0$.