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

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

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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

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

This is third degree Schur's inequality for \(pq,qr,rp\).

Hence proved. \(_\square\)

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

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

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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 :

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