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?

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

 A: 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. 
A: The inequality can be proven very cleanly using a powerful technique known as the Ravi Substitution. Before introducing this technique, let us first give a definition.

Definition An ordered triple $ (a,b,c) $ is said to be a triangular triple if and only if $ (a,b,c) \in \mathbb{R}_{+}^{3} $ and there exists a triangle whose edges have lengths $ a $, $ b $ and $ c $.

The following result says that triangular triples have a particularly nice form.

Theorem 1 An ordered triple $ (a,b,c) $ is a triangular triple if and only if there exists $ (x,y,z) \in \mathbb{R}_{+}^{3} $ such that
  $$
a = x + y, \quad b = y + z, \quad c = z + x.
$$
  Furthermore, $ x $, $ y $ and $ z $ are unique.

The Ravi Substitution is then the act of making the substitutions
$$
a \to x + y, \quad b \to y + z, \quad c \to z + x.
$$

We are now ready to prove the given inequality, so let $ (a,b,c) $ be a triangular triple. As
$$
\left| \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - \frac{a}{c} - \frac{b}{a} - \frac{c}{b} \right| < 1 \iff \left| a^{2} c + b^{2} a + c^{2} b - a^{2} b - b^{2} c - c^{2} a \right| < abc,
$$
it suffices to prove the inequality on the right.
Applying the Ravi Substitution, we obtain (after some algebraic manipulation, which I shall leave as an exercise)
$$
a^{2} c + b^{2} a + c^{2} b - a^{2} b - b^{2} c - c^{2} a = -(x - y)(y - z)(z - x).
$$
Hence,
\begin{align}
  &\left| a^{2} c + b^{2} a + c^{2} b - a^{2} b - b^{2} c - c^{2} a \right| \\
= &|-(x - y)(y - z)(z - x)| \\
= &|(x - y)(y - z)(z - x)| \\
= &|x - y||y - z||z - x| \\
< &(|x| + |y|)(|y| + |z|)(|z| + |x|) \quad (\text{By the Triangle Inequality.}) \\
= &(x + y)(y + z)(z + x) \quad (\text{As $ x,y,z > 0 $.}) \\
= &abc.
\end{align}

Further Notes on the Ravi Substitution
Given $ S \subseteq \mathbb{R}^{3} $, we call $ S $ a positive cone in $ \mathbb{R}^{3} $ if and only if $ S $ is closed under


*

*addition, i.e., $ \mathbf{x},\mathbf{y} \in S \implies \mathbf{x} + \mathbf{y} \in S $, and

*scalar multiplication by a positive real number, i.e., $ \mathbf{x} \in S, \lambda \in \mathbb{R}_{+} \implies \lambda \cdot \mathbf{x} \in S $.
Let $ \Delta $ denote the set of all triangular triples. Using the Triangle Inequality, it is easy to show that $ \Delta $ is a positive cone in $ \mathbb{R}^{3} $. Clearly, $ \mathbb{R}_{+}^{3} $ is also a positive cone in $ \mathbb{R}^{3} $. Next, define a mapping $ \mathcal{R}: \mathbb{R}_{+}^{3} \to \mathbb{R}^{3} $ as follows:
$$
\forall (x,y,z) \in \mathbb{R}_{+}^{3}: \quad \mathcal{R}(x,y,z) \stackrel{\text{def}}{=} (x + y,y + z,z + x).
$$

Theorem 2 The mapping $ \mathcal{R} $ is a bijective positively linear transformation from $ \mathbb{R}_{+}^{3} $ to $ \Delta $, i.e., $ \mathcal{R}: \mathbb{R}_{+}^{3} \to \Delta $ is a bijection and
  $$
\forall \mathbf{x},\mathbf{y} \in \mathbb{R}_{+}^{3}, ~ \forall \lambda \in \mathbb{R}_{+}: \quad \mathcal{R}(\lambda \cdot \mathbf{x} + \mathbf{y}) = \lambda \cdot \mathcal{R}(\mathbf{x}) + \mathcal{R}(\mathbf{y}).
$$
  We call $ \mathcal{R} $ the Ravi Substitution mapping.

Theorem 2 thus says that the Ravi Substitution mapping $ \mathcal{R} $ is an isomorphism of positive cones. Theorem 1 is an immediate consequence.
Proof of Theorem 2: Let $ (x,y,z) \in \mathbb{R}_{+}^{3} $, so $ \mathcal{R}(x,y,z) = (x + y,y + z,z + x) $. As
\begin{align}
(x + y) + (y + z) &= x + 2y + z > x + z, \\
(x + y) + (z + x) &= 2x + y + z > y + z, \quad \text{and} \\
(y + z) + (z + x) &= x + y + 2z > x + y,
\end{align}
there exists a triangle whose edges have lengths $ x + y $, $ y + z $ and $ z + x $. Hence, $ \text{Range}(\mathcal{R}) \subseteq \Delta $.
If $ (a,b,c) \in \Delta $, then by the Triangle Inequality, we have
$$
\left( \frac{a + c - b}{2},\frac{b + a - c}{2},\frac{c + b - a}{2} \right) \in \mathbb{R}_{+}^{3}.
$$
Hence, we can define a mapping $ \mathcal{S}: \Delta \to \mathbb{R}_{+}^{3} $ as follows:
$$
\forall (a,b,c) \in \Delta: \quad \mathcal{S}(a,b,c) \stackrel{\text{def}}{=} \left( \frac{a + c - b}{2},\frac{b + a - c}{2},\frac{c + b - a}{2} \right).
$$
As $ \mathcal{R} \circ \mathcal{S} = \text{id}_{\Delta} $ and $ \mathcal{S} \circ \mathcal{R} = \text{id}_{\mathbb{R}_{+}^{3}} $, we deduce that $ \mathcal{R}: \mathbb{R}_{+}^{3} \to \Delta $ is a bijection.
The proof that $ \mathcal{R} $ is a positively linear transformation is not difficult at all, so we leave it to the reader. $ \quad \spadesuit $
Suppose now that we have a function $ F: \Delta \to \mathbb{R} $ and are asked to prove
$$
\forall (a,b,c) \in \Delta: \quad F(a,b,c) \geq 0.
$$
By Theorem 2, this is equivalent to proving
$$
\forall (x,y,z) \in \mathbb{R}_{+}^{3}: \quad F(\mathcal{R}(x,y,z)) = F(x + y,y + z,z + x) \geq 0.
$$
The reason why the second statement may be easier to prove than the first one is that $ \mathbb{R}_{+}^{3} $ is easier to handle than $ \Delta $. (Do you find it easy to visualize $ \Delta $ in $ 3 $-space? I do not.) Many inequalities involving three variables, such as the three-variable version of the AM-GM Inequality, are tailor-made for $ \mathbb{R}_{+}^{3} $, so applying the Ravi Substitution mapping usually simplifies matters considerably as we may then use these inequalities directly.
