Inequality on the side lengths of a triangle: $\left| \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - \frac{a}{c} - \frac{b}{a} - \frac{c}{b} \right| < 1$. This problem is taken from the Kosovo Mathematical Olympiad for Grade-$ 10 $ students.
Let $ a $, $ b $ and $ c $ be the lengths of the edges of a given triangle.
How can one prove the following inequality?

$$
\left| \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - \frac{a}{c} - \frac{b}{a} - \frac{c}{b} \right| < 1.
$$


You can write the question as 
$|\frac{a^2c+b^2a+c^2b-b^2c-c^2a-a^2b}{abc}|<1$. This means you would like to show that 
$-1<\frac{a^2c+b^2a+c^2b-b^2c-c^2a-a^2b}{abc}<1$. Since $a,b,c$ are the sides of a triangle, you  have that $a,b,c>0$. This means you want to show $-abc<a^2c+b^2a+c^2b-b^2c-c^2a-a^2b<abc$. We write the inside term more compactly, we are trying to show $-abc<ac(a-c)+ab(b-a)+bc(c-b)<abc.$ Because the expression we are trying to show is cyclic, without loss of generality, we assume $a\geq b\geq c >0$. Since they are the sides of a triangle, you have that $a-b<c$, $b-c<a$ and $a-c<b$ (sum of two sides exceeds the third one). Therefore, this gives you $ac(a-c)<abc$. Also $a\geq b\geq c>0$ give you $ab(b-a)\leq 0$ and $bc(c-b)\leq 0$. Adding the three inequalities gives $ac(a-c)+ab(b-a)+bc(c-b)<abc$. 
Now since $b-c<a$ you have that $bc(b-c)<abc$. Call this (1). Since $b+c > a$ and since $b-c\geq 0$, it follows that $(b+c)(b-c)\geq a(b-c)$, which means $b^2-c^2-ab+ac\geq 0$ implying $b(a-b)+c(c-a)\leq 0$. This means that $ab(a-b)+ac(c-a)\leq 0$. Call this (2). 
Adding (1) and (2) side-by-side gives $ab(a-b)+ac(c-a)+bc(b-c)< abc$ which is the same as $-abc<ab(b-a)+ac(a-c)+bc(c-b)$, which is what we wanted to show. 
