How to solve the system $ax>y+z$, $by>x+z$, $cz>x+y$ in positive numbers?

Question

Let $a>b>c>1$. How to find solutions in positive numbers of the following system? \begin{cases} ax>y+z \\ by>x+z \\ cz>x+y \end{cases}

Answer 1

I assume $a,b,c$ are given and you want to find $x,y,z>0$ satisfying your system. In terms of $x$, we have

$$ \min(b y - z, cz - y) > x > y/a + z/a$$

In order for this interval for $x$ to be nonempty, we need

$$ \eqalign{(b - 1/a) y &> (1 + 1/a) z \cr (c - 1/a) z &> (1 + 1/a) y \cr} $$ So (noting that $c > 1 > 1/a$) $$ \dfrac{ab-1}{a+1} y > z > \dfrac{a+1}{ac-1} y $$

Finally, to make the interval for $z$ be nonempty, we need

$$ \dfrac{ab-1}{a+1} > \dfrac{a+1}{ac-1} $$

or equivalently

$$ a b c > a + b + c + 2 $$

and then any $y > 0$ will do, with $z$ and then $x$ in the intervals given above.

Answer 2

$$\begin{cases} ax>y+z \\ by>x+z \\ cz>x+y \end{cases}$$ So, $$ax+by+cz>2(x+y+z)$$ If it is an identity in x,y and z then this always holds for $a>b>c>2;\;x,y,z>0\wedge x,y,z\in\mathbb{R}$