# Cube of harmonic mean

Question:
The geometric mean of two numbers is $$8$$ while the arithmetic mean is $$4$$. Determine the cube of the harmonic mean. Answer is $$4096$$.

Can anyone tell me how to solve this problem? I do not know how since from what I've known, the AM of is always greater than GM. Please show me your complete solution

• Well, I believe the book forgot about the AM-GM inequality, and made an ill-posed problem. If the problem had made sense, the answer would have been $4096$. – Mankind Dec 20 '15 at 14:56
• This is impossible: the geometric mean cannot be more than the arithmetic mean. – Bernard Dec 20 '15 at 15:28
• I believe that the aim of the exercise was not to ignore constraints. AM-GM inequality assumes non-negative numbers. – user376343 May 26 '20 at 11:21

Since $\sqrt{ab}=8$ and $\frac{a+b}2=4$, we get that $a=4(1+i\sqrt3)$ and $b=4(1-i\sqrt3)$. Then $$\left[\frac2{\frac1{4(1+i\sqrt3)}+\frac1{4(1-i\sqrt3)}}\right]^{\,3}=16^3=4096$$ Since the GM is greater than the AM, the numbers cannot both be positive reals.

Of course, as was observed in a deleted answer $$\text{HM}=\frac2{\frac1a+\frac1b}=\frac{2ab}{a+b}=\frac{\text{GM}^2}{\text{AM}}$$

• Very Einstein-like, to spot the assumption that everyone else was making! – Paul Sinclair Dec 20 '15 at 20:00

The harmonic mean for $2$ numbers can be calculated as $$\frac{2}{\frac1a+\frac1b}=\frac2{\frac{a+b}{ab}}=\frac{2ab}{a+b}=ab\ \div \frac{a+b}{2}=\frac{\text{GM}^2}{\text{AM}}$$ So the harmonic mean is $8^2/4=16$. So your answer is $4096$.

Note: This is impossible as the GM cannot be more than AM.

This is very nice as an example of mindless problem composition, but if it was only from accidental switching of AM and GM when writing out the problem, then:

If AM = 8 and GM = 4, then the numbers are positive reals and HM=2.