Let $x,y,z$ be real numbers such that the numbers $$\frac{1}{|x^2+2yz|},\quad\frac{1}{|y^2+2zx|},\quad\frac{1}{|z^2+2xy|}$$are lengths of sides of a (non-degenerate) triangle. Determine all possible values of $xy+yz+zx$.
Problem
Source: Czech and Slovak Olympiad 2018, National Round, Problem 2
Tags: algebra, national olympiad
05.04.2018 19:30
$(x,y,z)=(t,t,t)$ and $(t,t,-2t)$ where $t\in \mathbb{R}^+$ gives us $xy+yz+zx$ can be all real numbers except zero. To show that $xy+yz+zx$ can't be zero, suppose to the contrary, then we get $\sum_{cyc}{\frac{1}{x^2+2yz}} =\sum_{cyc}{\frac{1}{(x-y)(x-z)}} =0$. Not hard to see that $|a|,|b|,|a+b|$ can't be side lengths of non-degenerate triangle for any $a,b\in \mathbb{R}$.
05.04.2018 19:33
Where you proved that "$xy+yz+zx$ can be all real numbers "?
05.04.2018 19:36
@above ThE-dArK-lOrD wrote: $(x,y,z)=(t,t,t)$ and $(t,t,-2t)$ where $t\in \mathbb{R}^+$
15.08.2018 12:00
I do not understand the above solution...
02.07.2024 21:08
soryn wrote: I do not understand the above solution... official solution: https://www.matematickaolympiada.cz/media/3459612/brozura_a67angl_new.pdf#page=13