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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.