# 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?

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

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What does "Problem grade 10" mean? –  Chris Eagle Jan 23 '13 at 17:20
@ChrisEagle The problem was given to 10th graders in Kosovo. –  Potato Jan 23 '13 at 17:28
I've edited the title. Seeing "Olympiad Mathematics Kosovo (Problem, Grade 10)" on the front page tells me absolutely nothing about the subject of the problem. –  Rahul Jan 25 '13 at 2:46

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.

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+1 Thanks for sharing. Nice technique. –  Sasha Jan 24 '13 at 3:29
@Sasha: You’re welcome! –  Haskell Curry Jan 24 '13 at 6:20
this technique is a cannon. and the question is obviously mosquitoes. thank you for sharing your knowledge with us. –  mathemagician Jan 25 '13 at 6:31
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.
I think you need the inequality to be symmetric to assume $a\ge b\ge c$. Being cyclic is insufficient. –  Christmas Bunny Mar 24 '13 at 4:35