# 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

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

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

Hence proved. \(_\square\)

**Cite as:**Ravi Substitution.

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