New Year Challenge: Find Real Number Triples

[anemone]

Find all triples $(a,\,b,\,c)$ of real numbers such that the following system holds:

$a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\\a^2+b^2+c^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}$

 

[jvicens]

To the original poster,

Thank you for your interesting question. my approach to solving this system would be to first simplify the equations by multiplying both sides by the common denominator $abc$. This would give us the following system:

$a^2bc+ab^2c+abc^2=1+bc+ac\\a^2b^2c^2+a^2bc^2+a^2b^2c=1+a^2c^2+b^2c^2$

Next, I would rearrange the equations to isolate one variable in terms of the other two. For example, in the first equation, we can isolate $c$ as follows:

$c=\dfrac{1+ab-a^2b}{ab+1-a^2}$

We can do the same for the other two variables in the second equation. This will give us three equations in terms of $a$, $b$, and $c$:

$c=\dfrac{1+ab-a^2b}{ab+1-a^2}\\a=\dfrac{1+bc-b^2c}{bc+1-b^2}\\b=\dfrac{1+ac-a^2c}{ac+1-a^2}$

From here, we can substitute these equations into each other and solve for one variable in terms of the other two. This will give us a quadratic equation, which can be solved to find the possible values for that variable. We can then substitute these values back into the original equations to find the corresponding values for the other two variables.

After solving for all three variables, we can check if the solutions satisfy the given equations. If they do, then they are valid solutions to the system. If not, then they are extraneous solutions and can be discarded.

I hope this helps in solving the system. Please let me know if you have any further questions. Good luck!