Ravi substitution in inequalities

Question

There is a well-known substitution for proving geometric inequalities: If $a,b,c$ are the side lengths of a triangle, then in an inequality involving $a,b,c$ it is possible to replace $a,b,c$ by $x+y,y+z,z+x$. This gets rid of the "triangle" restriction, so you just have to prove the resulting inequality in $x,y,z$ for any positive reals $x,y,z$.

My question concerns the other direction: what if you have an inequality in $x+y,y+z,z+x$, is it possible to make the substitutions $x+y=a$, $y+z=b$ and $z+x=c$ and end up with an inequality in $a,b,c$ that is valid for the side lengths of a triangle?

Yes. The existence of such positive $x,y,z$ is equivalent to $a,b,c$ being sides of a non-degenerate triangle. — Macavity, Mar 04 '15 at 14:37
Its easy to prove isn't it? Those satisfy the triangle inequality, and we are done. — Sawarnik, Mar 04 '15 at 16:47
@Macavity For the sake of moving this question off the unanswered list, would you mind posting your comment as an answer? — Peter Woolfitt, May 15 '15 at 02:20

score 2 · Accepted Answer · answered May 15 '15 at 05:02

2

Yes.

If you have an inequality in positive reals $x, y, z$, then by defining $a=x+y, b= y+z, c = z+x$, we get positive reals $a, b, c$ s.t. $a+b - c = 2y > 0, b+c-a=2z> 0, c+a-b=2x > 0$, so $a, b, c$ form a triangle.

answered May 15 '15 at 05:02

Macavity

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Ravi substitution in inequalities

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