Solving cyclic inequalities

I understand that in case of symmetric inequalities, you can assume abc a \geq b \geq c without loss of generality because you can replace for example aa with bb but still the inequality remains unchanged.

But in case of cyclic inequalities can we assume a=max(a,b,c) a = max(a,b,c) without loss of generality. If yes please explain the reason of WLOG in this case as well.

Example:- [RMO 2017 P6]

Let x,y,zx,y,z be real numbers, each greater than 11. Prove that x+1y+1+y+1z+1+z+1x+1x1y1+y1z1+z1x1\dfrac{x+1}{y+1}+\dfrac{y+1}{z+1}+\dfrac{z+1}{x+1} \leq \dfrac{x-1}{y-1}+\dfrac{y-1}{z-1}+\dfrac{z-1}{x-1}

The official solution uses a=max(a,b,c) a = max(a,b,c) .

#rmo #inequalities #wlog

Note by Santu Paul
8 months, 2 weeks ago

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Since the inequality (or equality) is cyclic, you can get new solutions by "rotating around" the numbers of a solution.

For example, if (5,2,9) (5,2,9) is a solution, then (2,9,5) (2,9,5) and (9,5,2) (9,5,2) are solutions as well, but (5,9,2) (5,9,2) isn't a solution.

This works because cyclic (in)equalities only use some operation (in your example a+1b+1 \frac {a+1}{b+1} for some a,b a,b ) of all pairs of two "consecutive" variables, but since all "consecutive" variable pairs ((x,y) (x,y) , (y,z) (y,z) and (z,x) (z,x) ) are used in the (in)equality, the order of the variables matters, but the "rotation" doesn't and that's why you can set a=max(a,b,c) a = \text{max} (a,b,c) .

Henry U - 8 months, 2 weeks ago

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Imagine all 3 (or any number of) variables arranged in a circle. The cyclic (in)equality involves some operation of all pairs of variables that are next to each other in this circle. You can't change the order of the elements in the circle and you also can't make a mirror image, but you can rotate the variables around and WLOG rotate one specific value (in your example the macimum value) so that it becomes a a .

Henry U - 8 months, 2 weeks ago

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I kind of get the geometric intuition but could you please be a bit more descriptive, especially in the algebraic one. Thank you.

SANTU PAUL - 8 months, 2 weeks ago

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If a cyclic equality has a solution (x,y,z)1=(a,b,c) (x,y,z)_1 = (a,b,c) , then – since the equality is cyclic – you can get another solution as (x,y,z)2=(b,c,a) (x,y,z)_2 = (b,c,a) . I think you might understand this intuitively, but to prove it algebraically we have to define what exactly a cyclic equality is.


My first thought is the definition

A cyclic equality is an equality f(x,y,z)=0 f(x,y,z) = 0 such that you can substitute (a,b,c)=(g(x,y),g(y,z),g(z,x)) (a,b,c) = (g(x,y),g(y,z),g(z,x)) for some g(r,s) g(r,s) into the equation and get a symmetric equation. In some cases you might have to make multiple substitutions.

Derived from the case of your inequality, we can show that

x+1y+1+y+1z+1+z+1x+1=0 \frac {x+1}{y+1} + \frac {y+1}{z+1} + \frac {z+1}{x+1} = 0

is cyclic (by my definition) because the function g(r,s)=r+1s+1 g(r,s) = \frac {r+1}{s+1} and its corresponding substitution (a,b,c)=(g(x,y),g(y,z),g(z,x)) (a,b,c) = (g(x,y),g(y,z),g(z,x)) gives the symmetric equality

a+b+c=0 a + b + c = 0 .

If we instead had the equality

xyyz+yzzx+zxxy=0 \frac {x-y}{y-z} + \frac {y-z}{z-x} + \frac {z-x}{x-y} = 0

then we would have to use the function g1(r,s)=rs g_1(r,s) = r-s

and get

ab+bc+ca=0 \frac ab + \frac bc + \frac ca = 0

Since this isn't symmetric, we have to use another substitution (d,e,f)=(g2(d,e),g2(e,f),g2(f,d)) (d,e,f) = (g_2(d,e),g_2(e,f),g_2(f,d)) for g2(r,s)=rs g_2(r,s) = \frac rs . This brings us to the symmetric equality

d+e+f=0 d + e + f = 0


Another possibility is to define a cyclic equality as an equality where we can get a new solution from a known one (x,y,z)1=(a,b,c) (x,y,z)_1 = (a,b,c) as (x,y,z)2=(b,c,a) (x,y,z)_2 = (b,c,a) .

Then, this fact you asked about is the defining property and therefore doesn't require any proof.


Which definition do you like the most, or do you have any other ideas?

Henry U - 8 months, 2 weeks ago

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