Maths Olympiad Prep

Library / /37 of 43

Algebra Difficulty 6.4 National Olympiad Prove it JBMO

Problem:
Let xi>1x_{i} > 1, for all i∈{1,2,3,…,2011}i \in \{1,2,3, \ldots, 2011\}. Prove the inequality
∑i=12011xi2xi+1−1≥8044 \sum_{i=1}^{2011} \frac{x_{i}^{2}}{x_{i+1}-1} \geq 8044
where x2012=x1x_{2012} = x_{1}. When does equality hold?

Solution

Solution:
Realize that (xi−2)2≥0⇔xi2≥4(xi−1)(x_{i}-2)^{2} \geq 0 \Leftrightarrow x_{i}^{2} \geq 4(x_{i}-1). So we get:
x12x2−1+x22x3−1+…+x20112x1−1≥4(x1−1x2−1+x2−1x3−1+…+x2011−1x1−1) \frac{x_{1}^{2}}{x_{2}-1} + \frac{x_{2}^{2}}{x_{3}-1} + \ldots + \frac{x_{2011}^{2}}{x_{1}-1} \geq 4\left(\frac{x_{1}-1}{x_{2}-1} + \frac{x_{2}-1}{x_{3}-1} + \ldots + \frac{x_{2011}-1}{x_{1}-1}\right)
By AM-GM:
x1−1x2−1+x2−1x3−1+…+x2011−1x1−1≥2011⋅x1−1x2−1⋅x2−1x3−1⋅…⋅x2011−1x1−12011=2011 \frac{x_{1}-1}{x_{2}-1} + \frac{x_{2}-1}{x_{3}-1} + \ldots + \frac{x_{2011}-1}{x_{1}-1} \geq 2011 \cdot \sqrt[2011]{\frac{x_{1}-1}{x_{2}-1} \cdot \frac{x_{2}-1}{x_{3}-1} \cdot \ldots \cdot \frac{x_{2011}-1}{x_{1}-1}} = 2011
Finally, we obtain that
x12x2−1+x22x3−1+…+x20112x1−1≥8044. \frac{x_{1}^{2}}{x_{2}-1} + \frac{x_{2}^{2}}{x_{3}-1} + \ldots + \frac{x_{2011}^{2}}{x_{1}-1} \geq 8044.
Equality holds when (xi−2)2=0, ∀i=1,…,2011(x_{i}-2)^{2} = 0,\ \forall i = 1, \ldots, 2011, or x1=x2=…=x2011=2x_{1} = x_{2} = \ldots = x_{2011} = 2.

All the denominators are greater than 00, so by Cauchy-Schwarz we have:
x12x2−1+x22x3−1+…+x20112x1−1≥(x1+x2+…+x2011)2x1+x2+…+x2011−2011 \frac{x_{1}^{2}}{x_{2}-1} + \frac{x_{2}^{2}}{x_{3}-1} + \ldots + \frac{x_{2011}^{2}}{x_{1}-1} \geq \frac{\left(x_{1} + x_{2} + \ldots + x_{2011}\right)^{2}}{x_{1} + x_{2} + \ldots + x_{2011} - 2011}
It remains to prove that
(x1+x2+…+x2011)2x1+x2+…+x2011−2011≥8044 \frac{\left(x_{1} + x_{2} + \ldots + x_{2011}\right)^{2}}{x_{1} + x_{2} + \ldots + x_{2011} - 2011} \geq 8044
or
(∑i=12011xi)2+4⋅20112≥4⋅2011⋅∑i=12011xi \left(\sum_{i=1}^{2011} x_{i}\right)^{2} + 4 \cdot 2011^{2} \geq 4 \cdot 2011 \cdot \sum_{i=1}^{2011} x_{i}
which is obviously true by AM-GM for (∑i=12011xi)2\left(\sum_{i=1}^{2011} x_{i}\right)^{2} and 4⋅201124 \cdot 2011^{2}.

Equality holds when x1+x2+…+x2011=4022x_{1} + x_{2} + \ldots + x_{2011} = 4022 and
x1x2−1=x2x3−1=…=x2011x1−1 \frac{x_{1}}{x_{2}-1} = \frac{x_{2}}{x_{3}-1} = \ldots = \frac{x_{2011}}{x_{1}-1}
or xi2−xi=xi−1xi+1−xi−1, ∀i=1,…,2011⇒∑i=12011xi2=∑i=12011xixi+2x_{i}^{2} - x_{i} = x_{i-1} x_{i+1} - x_{i-1},\ \forall i = 1, \ldots, 2011 \Rightarrow \sum_{i=1}^{2011} x_{i}^{2} = \sum_{i=1}^{2011} x_{i} x_{i+2} where x2012=x1x_{2012} = x_{1} and x2013=x2x_{2013} = x_{2}. This means that x1=x2=…=x2011x_{1} = x_{2} = \ldots = x_{2011}.
So equality holds when x1=x2=…=x2011=2x_{1} = x_{2} = \ldots = x_{2011} = 2 since x1+x2+…+x2011=4022x_{1} + x_{2} + \ldots + x_{2011} = 4022.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.