Does there exists any $x$ such that $x\geq AM\geq GM?$ If not,how do I prove it is unbounded?

Question

We know $AM\geq GM$.Or,in words,$AM$ has a minimum value when it is equal to $GM$?But,by any chance is there any way to find $x$ such that it is the upper limit of $AM$?

My inspiration for this question-

My friend gave me this question-

The volume of a cuboid=$64 m^3$.What is its maximum possible Total Surface Area?

My attempt-

By applying AM-GM I get,

$$\frac{lb+bh+lh}{3}\geq lb\cdot bh\cdot lh^{\frac13}$$

$$\implies2(lb+bh+lh)\geq 96$$

Thus,I get the lower limit of surface area=$96$.But,how do I find the upper limit of surface area?(or in other words how do I find the upper limit of the AM-GM inequality.)

Thanks for any help!!

Answer 1

In every such problem, I would like to suggest a good approach which does have a rigorous proof. The minimum and maximum of functions occur at either the boundaries or the symmetry point. In your case, at the symmetry point, the minimum occurs. Hence, the maximum is infinity.

How infinity? Consider a cuboid of infinite length and infinitesimal breadth and height.