Note that, by the cosine law, we can construct a triangle with sides $a,b,c$. Hence the other inequalities also follow from the cosine law.

Gryphos wrote: Note that, by the cosine law, we can construct a triangle with sides $a,b,c$. Hence the other inequalities also follow from the cosine law. Note that nobody told that $a$, $b$, $c$ are positive. It's easy to correct but it need some work

bigant146 wrote: Given real numbers $a$, $b$, $c$ satisfy inequality $\left| \frac{a^2+b^2-c^2}{ab} \right|<2$. Prove that they also satisfy equalities $\left| \frac{b^2+c^2-a^2}{bc} \right|<2$ and $\left| \frac{c^2+a^2-b^2}{ca} \right| <2$. The first relationship implies that $|a|$, $|b|$ and $|c|$ are the side of a triangle...the rest is straightforward by cosinus law

bigant146 wrote: Given real numbers $a$, $b$, $c$ satisfy inequality $\left| \frac{a^2+b^2-c^2}{ab} \right|<2$. Prove that they also satisfy equalities $\left| \frac{b^2+c^2-a^2}{bc} \right|<2$ and $\left| \frac{c^2+a^2-b^2}{ca} \right| <2$. See the solution 1 from here: https://www.cut-the-knot.org/m/Geometry/TriangleProblemFromCaucasus.shtml

The problems of Caucasus 2017 competition can be found here in English.

Attachments:
caucasus-17_eng.pdf (178kb)