The Buffalo Way

Imgur

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.

Note by Daniel Liu
5 years, 3 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

2lazy2bash

- 5 years, 3 months ago

#?

- 5 years, 3 months ago

Yes #2lazy2bash

- 5 years, 3 months ago

Are there any Olympiad problems that can be solved with Buffalo?

- 5 years, 3 months ago

Well, try solving the second one without buffalo's.

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.

- 5 years, 3 months ago

"You might even be docked a point on the USA(J)MO if you used buffalo way, because of it's extreme inelegantness."

You might actually get a score of a point on that USA(J)MO (correct answer, bad proof)

- 5 years, 3 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.

- 5 years, 3 months ago

We just have to make sure that all of our implications are reversible when using this method.

- 5 years, 3 months ago