What are the minimum and maximum values here? Let $\alpha, \beta, \gamma>0 $ and $\mu=\big(\alpha+\frac{1}{\alpha}\big)^{{\big(\beta+\frac{1}{\beta}\big)}^{\big(\gamma+\frac{1}{\gamma}\big)}}$
What are the minimum and the maximum values of $\mu$?
$\text{My attempt}$:
Since the sum of a positive number and its reciprocal is always at least two, the minimum value of $\mu$ is $2^{2^2}=2^4=16$, and there is no maximum value for $\mu$.
Am I correct?
 A: The function $f: \mathbb{R}_+ \rightarrow \mathbb{R}, x\mapsto f(x)=x+1/x$ is continuously differentiable for all $x$ in its domain. Its derivative is
$$f'(x)=1-1/x^2.$$
Thus an extremum $x^*$ is determined by the condition $f(x^*)=1-1/(x^*)^2=0$, such that $x^*=0$ is the unique extremum. As $f'(x)<0$ for $0<x<1$ and $f'(x)>0$ for $x>1$, the extremum is a minimum. The function $f$ evaluates to $f(x^*)=2$ at its minimum.
As a consequence, the original $\mu$ asked for assumes a minimum value where all the three expressions $f(x)$ are minimal for $x\in\{\alpha, \beta,\gamma\}$ and thus evaluates to $\mu_\textsf{min}=2^{(2^2)}=16$ as stated in the task description.
Moreover, $\mu$ is unbounded and can grow arbitrarily large. Thus a maximum does not exist on the unbounded domain. To prove this statement, assume this were not the case and $\mu\leq M$ for some fixed $M\in\mathbb{R}$ with $M>2$. Then take $\alpha>M$ and $\beta,\gamma$ arbitrary positive. As $f(x)\geq 2$ for all positive $x$, we have
$$\mu=M^{(2^2)}>M$$ because $M>1$ and $2^2>1$, a contradiction to our assumption.
A: This is an immediate consequence of AM-GM.
First, $u + 1/u \geq 2$ whenever $u > 0$.
Lower bound: by inequality above,
$\mu \geq {2^2}^2 = 16$, and this is attained with $\alpha = \beta = \gamma = 1$.
Upper bound: Using AM-GM again,
$
\mu \geq 4^{\gamma + 1/\gamma} \geq 4^{\gamma}. 
$
Therefore, $\sup_{\alpha, \beta,\gamma > 0} \mu = +\infty$,  while $\inf_{\alpha, \beta, \gamma > 0} \mu = 16$.
