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Tuesday, January 28, 2014

Four positive numbers

Let $a_1,a_2,a_3,a_4$ be four positive numbers and let: $$S_1 = a_1 + a_2 + a_3 + a_4$$ $$S_2 = \sum_{i \neq j} a_ia_j$$ $$S_3 = a_1a_2a_3 + a_1a_2a_4 + a_1a_3a_4 + a_2a_3a_4$$ $$S_4 = a_1a_2a_3a_4$$ Given that $$| \frac{S_1-S_3}{1-S_2+S_4} | < 1$$ show that there are two distinct $a_i,a_j$ such that : $$|a_i-a_j| < (\sqrt{2}-1)(1+a_ia_j)$$

Solution 1

If any two $a_i$ are the same then we are done. WLOG, we may now assume that $a_1 < a_2 < a_3 < a_4$. Suppose that for each two $a_i > a_j$ we always have: $$a_i - a_j > (\sqrt{2}-1)(1+a_ia_j)$$ Consider $a_3$ and $a_4$: $$(a_4 - a_3) > (\sqrt{2}-1)(1+a_4a_3)$$ $$(a_4 - a_3)(\sqrt{2}+1) > (1+a_4a_3)$$ $$a_4(\sqrt{2}+1 - a_3) > 1+a_3(\sqrt{2}+1)$$ Because the RHS is positive, in order for the LHS to be positive, we must have $a_3 < \sqrt{2} + 1$. Thus now we have $a_1 < a_2 < a_3 < \sqrt{2}+1$. Next, note that the function $f(x) = \frac{x + \sqrt{2} - 1}{1 - (\sqrt{2}-1)x}$ defined for $0 < x < \sqrt{2}+1$ is definite positive and strictly increasing. The assumption $$a_i - a_j > (\sqrt{2}-1)(1+a_ia_j)$$ for $j=1,2,3$ becomes: $$a_i(1 - (\sqrt{2}-1)a_j) > a_j + \sqrt{2}-1$$ And because $(1 - (\sqrt{2}-1)a_j) > 0 \iff a_j < \sqrt{2}+1$ we can divide both sides to obtain: $$a_i > \frac{a_i + \sqrt{2}-1}{1 - (\sqrt{2}-1)a_j} =f(a_j)$$ So now we have: $$a_4 > f(a_3), a_3 > f(a_2), a_2 > f(a_1)$$ We note that $f(0) = \sqrt{2}-1,f(\sqrt{2}-1) = 1, f(1) = \sqrt{2}+1$. Furthermore, because $f$ is strictly increasing, $$a_2 > f(a_1) > f(0) = \sqrt{2}-1$$ $$a_3 > f(a_2) > 1$$ $$a_4 > f(a_3) > \sqrt{2}+1$$ Also, if $a_2 \geq 1$ then we'd have $a_3 > f(a_2) > f(1) = \sqrt{2}+1$ a contradiction, so we must have $a_2 < 1$. By the same token, if $a_1 \geq \sqrt{2}-1$ we'd have $a_2 > f(a_1) > f(\sqrt{2}-1) = 1$ a contradiction, so we must have $a_1 < \sqrt{2} -1 $. Now we've established that: $$0 < a_1 < \sqrt{2} - 1 < a_2 < 1 < a_3 < \sqrt{2} + 1 < a_4$$ From there it's clear that $a_1a_3 < 1$ and $a_2a_4 > 1$. Next, we assert the following: $$a_1 + a_3 + a_1a_3 > 1$$ which is very easy to see from the bounds above. Also because $a_4 > 0$ and $a_2 < 1$, $$\frac{1}{a_2a_4} + \frac{1}{a_2} + \frac{1}{a_4} > 1$$ $$1 + a_2 + a_4 > a_2a_4$$ Therefore: $$(1-a_1a_3)(1-a_2a_4) > (a_1+a_3)(a_2+a_4)$$

Wednesday, January 15, 2014

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.