# Solve equation: $\log_2 \left(1+ \frac{1}{a}\right) + \log_2 \left(1 +\frac{1}{b}\right)+ \log_2 \left(1 + \frac{1}{c}\right) = 2$

$$\log_2 \left(1 + \frac{1}{a}\right) + \log_2 \left(1 + \frac{1}{b}\right)+ \log_2 \left(1 + \frac{1}{c}\right) = 2 \quad \text{where a, b, c \in N.}$$

Apparently, the answer is $$a= 1$$, $$b =2$$, and $$c\space = 3$$.

When I asked my math teacher I was told that the solution involved a bit of number theory, but didn't recieve a complete explanation. Could someone clear that up for me?

$$\log_2 \left(a + \frac{1}{a}\right) + \log_2 \left(b + \frac{1}{b}\right)+ \log_2 \left(c + \frac{1}{c}\right) = 2 \quad \text{where a, b, c \in N.}$$

My apologies for causing confusion.

• Unless I am mistaken, the equation does not hold for $(a, b, c) = (1, 2, 3)$. Perhaps it is $\log_2 (1 + \frac{1}{a}) + \log_2 (1 + \frac{1}{b})+ \log_2 (1 + \frac{1}{c}) = 2$? In that case you can find a solution on AoPS: artofproblemsolving.com/community/c6h43200p273208. May 24, 2020 at 10:04
• Yes, Indeed it is, I made a mistake. Thanks for pointing it out and sharing the link! May 24, 2020 at 10:56
• After your edit, you put Ishan's answer, with a decent amount of upvotes, in an unfavorable light. Also, the title still points at the old version. May 24, 2020 at 10:57
• @rtybase: On the other hand, if a question states that “apparently, the answer is ...” if it isn't then a request for clarification might be more appropriate than answering the (apparently wrong) question. May 24, 2020 at 11:11
• @MartinR indeed, I was just highlighting the fact. I hope guys will find a way to sort this little problem out. I'd simply add notes in the question and answer mentioning the edit (and time of the edit) and the change in the meaning of the question. Or close this one and create another ... May 24, 2020 at 11:17

Simplifying the LHS, we get $$(a^2+1)(b^2+1)(c^2+1) = 4abc$$ But, by the AM-GM inequality, we get $$x^2+1\ge2x$$, which gives $$(a^2+1)(b^2+1)(c^2+1) \ge 8abc$$