Infimum of abc

Find the largest number $M$ such that for any given distinct number $a,b,c$ such that $0 < a,b,c < 1$ we always have: $$\frac{(a+b)(b+c)(c+a)}{|(a-b)(b-c)(c-a)|} > M$$

Solution

Clearly $M=1$ satisfies the condition. Now we prove that we can get the LHS arbitrarily close to 1.

Let $a$ be an arbitrary number < 1, $b = a/(1+x)$ and $c=a/(1+x+yx)$ for some really large numbers $x,y$. The LHS now becomes: $$\frac{(2+x)(2+(y+1)x)(2+(y+2)x)}{y(y+1)x^3}$$ $$=(\frac{2}{x}+1)(\frac{2}{x(y+1)} + 1)(\frac{2}{xy} + \frac{2}{y} + 1)$$ We can set $x,y$ large enough that the expression is as close as needed to 1.

Polynomial Algorithm

Credit for this problem goes to Zeke Chin

Does there exist an algorithm for the decision problem:

Given a polynomial over $n$ variables $x_1,\cdots,x_n$ with positive integer coefficients, and a positive integer $c$, is it true that $cx_1\cdots x_n \leq P(x_1,\cdots,x_n)$ for all assignments of positive integers to $x_1,\cdots,x_n$?

For example, for $P = x_1^2 + x_2^2$ and $c=2$, the algorithm should return "Yes" since $2x_1x_2 \leq x_1^2 + x_2^2$

KBB2 Problem 4

For $a,b,c$ positive reals, show that:

$\displaystyle \frac{6a}{9a^2+5(a+b+c)^2} + \frac{6b}{9b^2+5(a+b+c)^2} + \frac{6c}{9c^2+5(a+b+c)^2} \leq \frac{1}{a+b+c}$