The **Buffalo Way** is a plug-and-bash method used to solve olympiad inequalities. It is usually applied to symmetric inequalities, where we can assume WLOG that the variables are in a specific order; that is, \(x_1 \le x_2\le \cdots \le x_n\).

To illustrate this method, we shall prove \(AM-GM\) for two variables using the method.

Prove that \[\dfrac{x+y}{2}\ge \sqrt{xy}\] for non-negative reals \(x,y\).

First, assume WLOG that \(x\le y\). Thus, we can represent \(x\) and \(y\) by: \[x=a\] \[y=a+b\] where \(a,b\) are non-negative reals. Make sure you see why this is true.

Thus, we want to prove \[\dfrac{a+a+b}{2}\ge \sqrt{a(a+b)}\implies a+\dfrac{b}{2}\ge \sqrt{a^2+ab}\]

Squaring both sides gives \[a^2+ab+\dfrac{b^2}{4}\ge a^2+ab\implies \dfrac{b^2}{4}\ge 0\] which is true by the trivial inequality.

In general, if we have variables satisfying \(x_1 \le x_2\le \cdots \le x_n\), then we substitute \[x_1=y_1\]\[x_2=y_1+y_2\] \[\vdots\] \[x_n=y_1+y_2+\cdots +y_n\]

Given that \(a,b,c\) are non-negative reals such that \(a\le b\le c\), then prove that \[(a+b)(c+a)^2\ge 6abc\]

We see that we already have the condition of \(a\le b\le c\), so we can apply Buffalo's Way directly. Let \[a=x\] \[b=x+y\] \[c=x+y+z\] where \(x,y,z\) are non-negative reals. Thus, we want to prove \[(2x+y)(2x+y+z)^2\ge 6x(x+y)(x+y+z)\]

This expands to (told you Buffalo Way is a bash): \[8x^3+12x^2y+8x^2z+6xy^2+8xyz+2xz^2+y^3+2y^2x+yz^2\ge 6x^3+12x^2y+6x^2z+6xy^2+6xyz\]

Rearranging gives \[2x^3+2x^2z+2xyz+2xz^2+y^3+2y^2z+yz^2\ge 0\] which is trivially true (\(x,y,z\) are positive).

We can also see that this method gives an equality case: We must have \(x=y=z=0\). Thus, \(a=b=c=0\).

When actually solving Olympiad Inequalities in competitions, **NEVER** use this method, unless you have no idea now to do it otherwise. In cases where the inequality is relatively simple (no denominators) the Buffalo Way is almost guaranteed to work.

NOTE: I don't know where the name comes from. If you try to search it up, you won't get any results. I first heard of this method from the AoPS Community.

## Comments

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TopNewest## 2lazy2bash

– David Lee · 2 years, 4 months agoLog in to reply

`#`

? – Ahaan Rungta · 2 years, 4 months agoLog in to reply

– David Lee · 2 years, 3 months ago

Yes #2lazy2bashLog in to reply

We just have to make sure that all of our implications are reversible when using this method. – Bob Krueger · 2 years, 4 months ago

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Are there any Olympiad problems that can be solved with Buffalo? – Kaan Dokmeci · 2 years, 4 months ago

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But to answer your question, chances are there are no Olympiad question that can be only exclusively solved using the Buffalo way. This method is the least elegant method of olympiad inequalities ever, and it is almost guaranteed that there is a more elegant solution.

You might even be docked a point on the USA(J)MO if you used buffalo way, because of it's extreme inelegantness. – Daniel Liu · 2 years, 4 months ago

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You might actually get a score of a point on that USA(J)MO (correct answer, bad proof) – David Lee · 2 years, 4 months ago

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– Daniel Liu · 2 years, 4 months ago

No, if you have a complete proof, no matter how inelegant, then you will get at least 5-6 points. It's impossible to get 1 point if you write a complete proof.Log in to reply