I know it is based on the property that f (x,y,z)=t^(-n)f (tx,ty,tz). So any inequality can be written as a function of several variables and this property can be applied if the function is homogeneous. I would like to know it's various applications.

The basic idea is that a homogeneous polynomial (IE polynomial whose terms all have the same degree) can be easier to deal with than a polynomial whose terms have different degrees. As such, if we're given a condition, we try and use that to make the terms equal.

For example, a common condition is of the form \( x + y + z = 1 \) (or some other constant). If we had to deal with the expression \( xy+yz+zx - xyz \), we can homogenize by multplying the degree 2 terms by \( x+y+z = 1 \), giving us

@Calvin Lin
–
I have seen problems where no such constraints as above is given but one assumes such a constraint then solves the problem. How is this possible?

@Abhi Kumbale
–
In those cases, an assumption like \( x + y + z = t \) is made, and then we introduce another variable.

In a sense, it is similar to "Richard Feymanns favourite integration trick of differentiating through the integral" (assuming you are familiar with how that works).

The given expression is homogeneous equation of degree \(0\). So, we can make a constraint \(a+b+c+d=1\).

Do you understand why that means?

Suppose that \( a + b + c + d = t \), then consider the change of variables \( A = \frac{a}{t} ,\) etc. What is \( A + B + C + D \)?

Why does "homogenous equation of degree 0" allow us to do this substitution? What happens if we have a "homogenous equation of degree 1"? What happens for a general polynomial say \( a^2b + cd + c + 2 \)?

Sorry for being unable to answer due to the fact I have no in depth knowledge of inequalities.
But, I advise you to read Calvin Sir's note on 'How to ask for help?'; unable to provide link due to an error.
It is in no way to offend you and I apologise if it does.

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TopNewestWhat do you currently understand about it?

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I know it is based on the property that f (x,y,z)=t^(-n)f (tx,ty,tz). So any inequality can be written as a function of several variables and this property can be applied if the function is homogeneous. I would like to know it's various applications.

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The basic idea is that a homogeneous polynomial (IE polynomial whose terms all have the same degree)

can be easier to deal withthan a polynomial whose terms have different degrees. As such, if we're given a condition, we try and use that to make the terms equal.For example, a common condition is of the form \( x + y + z = 1 \) (or some other constant). If we had to deal with the expression \( xy+yz+zx - xyz \), we can homogenize by multplying the degree 2 terms by \( x+y+z = 1 \), giving us

\[ xy+yz+zx - xyz \\ = (x+y+z)(xy+yz+zx) - xyz \\ = (x+y)(y+z)(z+x) \\ = (1-x)(1-y)(1-z) \\ \leq \left( \frac{2}{3} \right)^3 \]

This is a (constructed) example where the homogenization leads to a nice factoring, which leads to a nice way to deal with the expression.

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In a sense, it is similar to "Richard Feymanns favourite integration trick of differentiating through the integral" (assuming you are familiar with how that works).

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https://brilliant.org/profile/surya-ou9y8q/sets/olympiad-proof-problems/354510/olympiad-proof-problem-day-1/

How is this right? I have also seen a similar approach in Engel's book.

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Do you understand why that means?

Suppose that \( a + b + c + d = t \), then consider the change of variables \( A = \frac{a}{t} ,\) etc. What is \( A + B + C + D \)?

Why does "homogenous equation of degree 0" allow us to do this substitution? What happens if we have a "homogenous equation of degree 1"? What happens for a general polynomial say \( a^2b + cd + c + 2 \)?

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Sorry for being unable to answer due to the fact I have no in depth knowledge of inequalities.

But, I advise you to read Calvin Sir's note on 'How to ask for help?'; unable to provide link due to an error. It is in no way to offend you and I apologise if it does.

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@Calvin Lin @Surya Prakash @Aditya Raut @Mark Henning @Sharky Kesa

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