Question: Let a, b, c be three natural numbers. If their arithmetic mean is less than the harmonic mean of $2, 3, 4$ by $\frac{10}{13}$, then find the value of $abc$.
I started the question by making the following equation: $$\frac{a+b+c}{3} = \frac{36}{13} - \frac{10}{13}$$ and then realized that this equation can't get me far as I need the three terms, $a,b,c$ to be in multiplication. What should I do?
The harmonic mean of $2,3,4$ is given by $$ \frac3{\frac12+\frac13+\frac14}=\frac3{\frac6{12}+\frac4{12}+\frac3{12}}=\frac{36}{13}. $$ We have that $$ \frac{a+b+c}3=\frac{36}{13}-\frac{10}{13}=2. $$ By the inequality of arithmetic and geometric means, $$ \sqrt[3]{abc}\le2 $$ or $$ abc\le 8. $$ So we know that $abc\le8$ and $a+b+c=6$. Hence, there are a a few possible values of $abc$ that meet the criteria. Suppose that $a=4$, $b=c=1$, then $abc=4$. If $a=b=c=2$, then $abc=8$.