Difference between revisions of "2011 JBMO Problems/Problem 1"

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==See Also==
 
==See Also==
{{JBMO box|year=2011|before=First Problem|num-a=2}}
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Latest revision as of 17:32, 12 April 2019

Problem

Let $a,b,c$ be positive real numbers such that $abc = 1$. Prove that:

$\prod(a^5+a^4+a^3+a^2+a+1)\geq 8(a^2+a+1)(b^2+b+1)(c^2+c+1).$

Solution

Since $abc = 1$, $a = \tfrac{1}{bc}$. By AM-GM Inequality, \begin{align*} b^3c^3 + \frac{1}{b^3c^3} &\ge 2 \\ b^3 + \frac{1}{b^3} &\ge 2 \\ c^3 + \frac{1}{c^3} &\ge 2. \end{align*} Adding the inequalities together and factoring yields \begin{align*} 1 + c^3 + \frac{1}{c^3} + 1 + b^3 + b^3c^3 + \frac{1}{b^3c^3} + \frac{1}{b^3} &\ge 8 \\ (1+c^3)(1 + \frac{1}{c^3} + b^3 + \frac{1}{b^3c^3}) &\ge 8 \\ (1+c^3)(1+b^3)(1 + \frac{1}{b^3c^3}) &\ge 8 \\ (1+c^3)(1+b^3)(1 + a^3) &\ge 8. \end{align*} Since $a,b,c$ are positive, $a^2 + a + 1$, $b^2 + b + 1$, and $c^2 + c + 1$ are all greater than 0. That means \begin{align*} \prod(a^3 + 1)(a^2 + a + 1) &\ge 8(a^2+a+1)(b^2+b+1)(c^2+c+1) \\ \prod \left( (a^3 + 1)\frac{a^3 - 1}{a-1} \right) &\ge 8(a^2+a+1)(b^2+b+1)(c^2+c+1) \\ \prod \frac{a^6 -1}{a-1} &\ge 8(a^2+a+1)(b^2+b+1)(c^2+c+1) \\ \prod(a^5+a^4+a^3+a^2+a+1) &\ge 8(a^2+a+1)(b^2+b+1)(c^2+c+1). \end{align*}

See Also

2011 JBMO (ProblemsResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4
All JBMO Problems and Solutions
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