USA IMO

First Solution. Assume, for the sake of contradiction, that at least two of the numbers $$\frac{2}{a}+\frac{3}{b}+\frac{6}{c},\frac{2}{b}+\frac{3}{c}+\frac{6}{a},\frac{2}{c}+\frac{3}{a}+\frac{6}{b}$$ are less than $6$. Without loss of generality, we may further assume that the first and the last are less than $6$. Then $$\frac{5}{a}+\frac{9}{b}+\frac{8}{c}<12.$$ Also, because $b+c\ge a(bc-1)$, we have $$\frac{1}{a}\ge \frac{bc-1}{b+c}.$$ It follows that $$\frac{5(bc-1)}{b+c}+\frac{9}{b}+\frac{8}{c}<12,$$ or $$5b^2{c}^2+12bc-12b^{2}c-12c^{2}b+9c^2+8b^2<0.(1)$$ Completing squares yields $$(2bc-2b-3c{)}^2+b^2(c-2{)}^2<0,$$ a contradiction. This proves the conclusion. To obtain equalities, that is two of the members $$\frac{2}{a}+\frac{3}{b}+\frac{6}{c},\frac{2}{b}+\frac{3}{c}+\frac{6}{a},\frac{2}{c}+\frac{3}{a}+\frac{6}{b}$$ are equal to $6$, we must have $c-2=0$ and $2bc-2b-3c=0$, that is $c=2$, $b=3$, $a=1$. Therefore the equalities hold if and only if $(a,b,c)$ is one of the triples $(1,3,2)$, $(3,2,1)$, $(2,1,3)$.

Or: Rewriting (1) as a quadratic form in $b$ yields $$(5c^2-12c+8)b^2-12c(c-1)b+9c^2<0,$$ which is impossible, because the leading coefficient $$5c^2-12c+8=5{\left(c-\frac{6}{5}\right)}^2+\frac{4}{5}$$ is always positive and the discriminant $$\begin{array}{rl}\Delta & =[12c(c-1){]}^2-36c^2(5c^2-12c+8)\\ & =36c^2[4(c-1{)}^2-(5c^2-12c+8)]=-36c^2(c-2{)}^2\end{array}$$ is always nonpositive.

Second Solution. (by David Shin) Perform the substitutions $x=1/a$, $y=1/b$, and $z=1/c$. It suffices to prove that at least two of the inequalities $$2x+3y+6z\le 6,2y+3z+6x\le 6,2z+3x+6y\le 6$$ are true, where $$x,y,z>0\text{and}xy+yz+zx\ge 1.$$ Assume, for the sake of contradiction, that at least two of the given inequalities are false. Without loss of generality, we may assume that $2x+3y+6z<6$ and $2y+3z+6x<6$. Then $$\begin{array}{rl}& 144>[(2x+3y+6z)+(2y+3z+6x){]}^2\\ & =(8x+5y+9z{)}^2\\ & =64x^2+80xy+25y^2+81z^2+90yz+144zx\\ & =64x^2-64xy+16y^2+9y^2-54yz+81z^2+144(xy+yz+zx)\\ & =(8x-4y{)}^2+(3y-9z{)}^2+144\ge 144,\end{array}$$ a contradiction. Thus, our assumptions is false and at least two of the desired inequalities must be true. To obtain equalities, that is two of the numbers $$\frac{2}{a}+\frac{3}{b}+\frac{6}{c},\frac{2}{b}+\frac{3}{c}+\frac{6}{a},\frac{2}{c}+\frac{3}{a}+\frac{6}{b}$$ are equal to $6$, we must have $8x-4y=3y-9z=0$, or $a:b:c=(1/x):(1/y):(1/z)=2:1:3$. Therefore the equalities hold if and only if $(a,b,c)=(1,3,2)$, $(3,2,1)$, $(2,1,3)$.