Cube of harmonic mean

Question

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

Answer 1

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 $$ \text{HM}=\frac2{\frac1a+\frac1b}=\frac{2ab}{a+b}=\frac{\text{GM}^2}{\text{AM}} $$

Answer 2

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.

Answer 3

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.