## Ravi Substitution explained

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 and only if if there exist three positive reals $x,y,z$ such that $a=x+y, b=y+z$ and $c=z+x$. For the proof of this, I will let $DEF$ be the intouch triangle of $ABC$, as shown in the diagram (click it to make it bigger).

By the definition of the incentre $I$ and incircle $(I)$, the sides of the triangle $ABC$ are tangent to $(I)$ at the points $D,E$ and $F$. By considering the power of the points $A,B,C$ with respect to $(I)$ clearly we have $AF=AE,BE=BD$ and $CD=CF$. Letting $AF=AE=z,$ $BE=BD=x$ and $CD=CF=y$ shows that:

$a=BC=BD+DC=x+y\\ b=CD=CF+FA=y+z\\ c=AB=AE+EB=z+x$

as desired. So we have proved if $a,b,c$ are sides of a triangle then there exist positive reals $x,y,z$ satisfying $a=x+y,b=y+z$ and $c=z+x$.

Now let us assume $a=x+y,b=y+z$ and $c=z+x$ where all variables are positive reals. We wish to show that $a,b,c$ are sides of a non-degenerate triangle. This is only true, recalling the triangle inequality, if $a+b>c$ and the analogues hold. But using the substitution, it is equivalent to $x+2y+z>x+z$ i.e. $2y>0$, which is true since $y$ is positive.

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

Let $a,b,c$ be positive reals. Show that:

$abc\ge (a+b-c)(b+c-a)(c+a-b)$.

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 $a+b>c,b+c>a$ and $c+a>b$ or in other words, $a,b,c$ are the sides of a triangle. It is here we prompt the Ravi Substitution: replace $(a,b,c)$ with $(x+y,y+z,z+x)$. The inequality then transforms to:

$(x+y)(y+z)(z+x)\ge 8xyz$

for the positive reals $x,y,z$. Indeed this is an inequality any beginner should have encountered: $x+y\ge 2\sqrt{xy}$ by AM-GM, and take the product of the analogues to prove the result.

Here’s another one:

Let $a,b,c$ be the sides of a triangle. Prove that

$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}<2$

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

$\dfrac{x+y}{x+y+2z}+\dfrac{y+z}{y+z+2x}+\dfrac{z+x}{z+x+2y}<2$

Clearly $\dfrac{x+y}{x+y+2z}<\dfrac{x+y}{x+y+z}$.

Hence $\sum\dfrac{x+y}{x+y+2z}<\sum\dfrac{x+y}{x+y+z}=\dfrac{2x+2y+2z}{x+y+z}=2$.

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

Let $a,b,c$ be the lengths of the sides of a triangle. Prove that

$\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}\le\sqrt{a}+\sqrt{b}+\sqrt{c}$

and determine when equality occurs.

Through Ravi Substitution, the inequality is equivalent to

$\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge\sqrt{2x}+\sqrt{2y}+\sqrt{yz}$

After squaring and simplifying (this is not as painful as it looks), it reduces to proving $\sum\sqrt{(x+y)(y+z)}\ge\sum 2\sqrt{xy}$ where the sum is cyclic.

In fact,

$\sqrt{(x+y)(y+z)} \ge\sqrt{xy}+\sqrt{yz}\\ \iff xy+yz+y^2+zx \ge xy+yx+2y\sqrt{zx}\\ \iff y^2+zx\ge 2y\sqrt{zx}$.

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 $a,b,c$ be the side-lengths of a triangle $ABC$ where $s,R,r$ have their usual definitions. Prove that

$a(s-a)+b(s-b)+c(s-c)\le 9Rr$

2) For $a,b,c$ the sides of some triangle, prove the inequality

$\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}+\dfrac{1}{a+b-c}\ge\dfrac1a+\dfrac1b+\dfrac1c$

3) Find the smallest constant $k$ such that

$k>\dfrac{a^2+b^2+c^2}{ab+bc+ca}$

where $a,b,c$ are the sides of a triangle.

4) Prove that if $a,b,c$ are the sides of a triangle then

$(-a+b+c)(a-b+c)+(a-b+c)(a+b-c)+(a+b-c)(-a+b+c)$

$\le\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})$

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

$(ab+bc+ca)(a+b+c)>a^3+b^3+c^3+5abc$

(Look for ways to avoid expanding completely).

6) Suppose that $a,b$ and $c$ are the sides of a triangle. Prove that

$a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)\le 3abc$

Note:

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

### 7 Responses to “Ravi Substitution explained”

1. LingBin Says:

I am a Chinenese middle school teacher.

2. marko Says:

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

3. marko Says:

Sorry,i meant 9

4. ak1024 Says:

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🙂

5. Drilon Says:

Thanks a lot, good explained🙂

6. Elisha Shon Says:

What is the origin of the name?