# Ravi Substitution

Ravi substitution is a technique emerging as extremely useful in the field of geometric inequalities, particularly at Olympiad level. Some inequalities with the variables $a,b,c$ have the side constraint that $a,b,c$ are the sides of a triangle, however it may seem unclear how to use this to our advantage when solving a problem.

Ravi substitution offers a solution to this: $a,b,c$ are the sides of a non-degenerate triangle if there exist three positive reals $x,y,z$ such that $a = x+y$, $b = y+z$ and $c = z+x$.

The substitution is effective even in the most difficult problems involving the sides of a triangle. The substitution accompanied with clever manipulations and use of other basic inequalities gives beautiful solutions.

An example from IMO Shortlist 2006.

Problem (ISL 2006 A5):

If $a,b,c$ are the sides of a triangle, prove that

$\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3.$

Since $a,b,c$ are the sides of a triangle, we can set

$a=(q+r)^2,\quad b=(p+r)^2,\quad c=(p+q)^2.$

Thus, the inequality can be rewritten as

$\sum_\text{cyclic} \frac{\sqrt{2p^2+2pr+2pq-2qr}}{p} \le 6 .$

Thus, is it is sufficient to show that

$\begin{aligned} \sum_\text{cyclic} \frac{2p^2+2pr +2pq -2qr}{p^2} &\le 12\\ \iff \sum_\text{cyclic} \frac{pr+pq-qr}{p^2} &\le 3\\ \iff \sum_\text{cyclic} q^2r^2(pr+pq-qr) &\le 3p^2q^2r^2. \end{aligned}$

This is third degree Schur's inequality for $pq,qr,rp$.

Hence proved. $_\square$

## The complete Ravi substitution

A common way to use the Ravi substitution is by expressing x, y and z in terms of a, b and c. Meaning :

Since

$\left\{\begin{matrix} a = y + z \\ b = x + z \\ c = x+y\end{matrix}\right.$

Then

$\left\{\begin{matrix} x = \frac{b+c-a}{2} \\ y = \frac{a+c-b}{2} \\ z = \frac{a+b-c}{2}\end{matrix}\right.$

Memorizing the order of $a$ , $b$ and $c$ may be quite difficult, but usually there is no need since most of the inequalities requiring Ravi are symmetric. It's however possible to use a mnemonic method to keep it in mind :

In the first part, you can notice that " *$a$ doesn't have $x$ , $b$ doesn't have $y$ and $c$ doesn't have $z$* ".

In the bottom part, notice as well that " *$x$ has $-a$ , y has $-b$ and $z$ has $-c$* ".

Here is an application :

If $a,b,c$ are the sides of a triangle, prove that

$(a+b) (b+c) (c+a) \geqslant 8 (a+b-c) (b+c-a) (c+a-b)$

We can start from both sides, but doing it from the right is more Ravi-like :

According to the Ravi substitution, we can have $x, y, z$ such as :

$\left\{\begin{matrix} x = \frac{b+c-a}{2} \\ y = \frac{a+c-b}{2} \\ z = \frac{a+b-c}{2}\end{matrix}\right.$

We will also assume the reader knows AM-GM inequality

Therefore : $8 (a+b-c) (b+c-a) (c+a-b) = 8 \times 8 \frac{b+c-a}{2} \frac{a+c-b}{2} \frac{a+b-c}{2} = 8 \times 8 xyz = 8 \times 8 \sqrt{xy} \sqrt{yz} \sqrt{zx}$

AM-GM tells us that

$8 \times 8 \sqrt{xy} \sqrt{yz} \sqrt{zx} \leq 8 \times 8 \frac{x+y}{2} \frac{z+y}{2} \frac{x+z}{2}$

but

$8 \times 8 \frac{x+y}{2} \frac{z+y}{2} \frac{x+z}{2} = 8 abc = 8 \sqrt{ab} \sqrt{bc} \sqrt{ca}$

again, AM-GM tells us

$8 \sqrt{ab} \sqrt{bc} \sqrt{ca} \leq 8 \frac{a+b}{2} \frac{b+c}{2} \frac{c+a}{2}$

but

$8 \frac{a+b}{2} \frac{b+c}{2} \frac{c+a}{2} = (a+b) (b+c) (c+a)$ and thus we have :

$(a+b) (b+c) (c+a) \geqslant 8 (a+b-c) (b+c-a) (c+a-b)$

**Cite as:**Ravi Substitution.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/ravi-substitution/