Ravi Substitution is a technique emerging as extremely useful in the field of geometric inequalities, particularly at Olympiad level. Some inequalities with the variables have the side constraint that 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: are the sides of a non-degenerate triangle if and only if if there exist three positive reals such that and . For the proof of this, I will let be the intouch triangle of , as shown in the diagram (click it to make it bigger).

By the definition of the incentre and incircle , the sides of the triangle are tangent to at the points and . By considering the power of the points with respect to clearly we have and . Letting and shows that:

as desired. So we have proved if are sides of a triangle then there exist positive reals satisfying and .

Now let us assume and where all variables are positive reals. We wish to show that are sides of a non-degenerate triangle. This is only true, recalling the triangle inequality, if and the analogues hold. But using the substitution, it is equivalent to i.e. , which is true since is positive.

Next, let me show applications of Ravi Substitution. Consider the following simple example:

Let be positive reals. Show that:

.

Why does this inequality seem different from a typical, three-variable, Olympiad inequality? I’ll answer that for you. It’s the fact that we have minuses – the product on the right is not guaranteed to be positive. In fact at most one of the three terms on the right hand side can be negative, and this is easy to prove. But in this case, the RHS is negative whereas the LHS is clearly positive!

Hence we may assume none of the terms on the RHS is negative. This means and or in other words, are the sides of a triangle. It is here we prompt the Ravi Substitution: replace with . The inequality then transforms to:

for the positive reals . Indeed this is an inequality any beginner should have encountered: by AM-GM, and take the product of the analogues to prove the result.

Here’s another one:

Let be the sides of a triangle. Prove that

This time there is no preamble: after using the Ravi Substitution, we need to prove

Clearly .

Hence .

A little less trivial is the following problem from the 1996 Asian Pacific Maths Olympiad:

Let be the lengths of the sides of a triangle. Prove that

and determine when equality occurs.

Through Ravi Substitution, the inequality is equivalent to

After squaring and simplifying (this is not as painful as it looks), it reduces to proving where the sum is cyclic.

In fact,

.

This is true by AM-GM, so we are done.

I’ve complied problems to try for yourself, but remember there’s always multiple ways of proving an inequality; try to solve the problems both with and without Ravi Substitution. Good luck!

1) Let be the side-lengths of a triangle where have their usual definitions. Prove that

2) For the sides of some triangle, prove the inequality

3) Find the smallest constant such that

where are the sides of a triangle.

4) Prove that if are the sides of a triangle then

5) If are the sides of a triangle, show that

*(Look for ways to avoid expanding completely)*.

6) Suppose that and are the sides of a triangle. Prove that

*Note:*

A pdf version of this should be available when I have the time and motivation. Also, I will continue to append problems.

March 10, 2011 at 7:24 am |

I am a Chinenese middle school teacher.

November 13, 2011 at 11:47 am |

The 5th problem requires that the LHS is multiplied by 3 i think :)

November 13, 2011 at 11:52 am |

Sorry,i meant 9

November 13, 2011 at 1:21 pm |

Hi marko, you are right, for the 5th problem there was a transcription error. It has now been edited to the correct statement, sorry!

But in fact, multiplying the LHS by 3 would have sufficed, you were right the first time :)

June 26, 2013 at 5:06 am |

Thanks a lot, good explained :)