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

(Solution to Problem 1 -- neat inequality problem)
 
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\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).
 
\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*}</cmath>
 
\end{align*}</cmath>
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==Solution 2 (credit to dskull16)==
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Note that the inequality clearly holds when <math>a=b=c=1</math> and the method below can be used to verify the inequality is true in the case where one of <math>a,b,c</math> is one (to avoid complications with dividing by zero).
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Write <math>a^5+a^4+a^3+a^2+a+1 = \frac{a^6-1}{a-1}</math>
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Write <math>a^2 + a + 1 = \frac{a^3-1}{a-1}</math> by the sum of a geometric series formula. Thus we get:
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<math>\frac{a^6-1}{a-1}\cdot\frac{b^6-1}{b-1}\cdot\frac{c^6-1}{c-1} \geq 8\cdot\frac{a^3-1}{a-1}\cdot\frac{b^3-1}{b-1}\cdot\frac{c^3-1}{c-1}</math>
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Multiply both sides by <math>(a-1)(b-1)(c-1)</math> and recognise the difference of two squares to get:
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<math>(a^3+1)(b^3+1)(c^3+1) \geq 8</math>
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<math>\implies (abc)^3 + a^3 + b^3 + c^3 + (ab)^3 + (bc)^3 + (ac)^3 + 1 \geq 8</math>
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<math>\implies a^3 + b^3 + c^3 + (ab)^3 + (bc)^3 + (ac)^3 \geq 6</math>
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Now use AM-GM on each of <math>a^3 + b^3 + c^3</math> and <math>(ab)^3 + (bc)^3 + (ac)^3</math> to obtain:
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<math>a^3 + b^3 + c^3 \geq 3</math>
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<math>(ab)^3 + (bc)^3 + (ac)^3 \geq 3</math>
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which combines to give the final result.
  
 
==See Also==
 
==See Also==
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{{JBMO box|year=2011|before=First Problem|num-a=2}}
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{{JBMO box|year=2011|before=First Problem|num-a=2|five=}}
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[[Category:Intermediate Algebra Problems]]

Latest revision as of 17:00, 3 March 2024

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*}

Solution 2 (credit to dskull16)

Note that the inequality clearly holds when $a=b=c=1$ and the method below can be used to verify the inequality is true in the case where one of $a,b,c$ is one (to avoid complications with dividing by zero).

Write $a^5+a^4+a^3+a^2+a+1 = \frac{a^6-1}{a-1}$

Write $a^2 + a + 1 = \frac{a^3-1}{a-1}$ by the sum of a geometric series formula. Thus we get:

$\frac{a^6-1}{a-1}\cdot\frac{b^6-1}{b-1}\cdot\frac{c^6-1}{c-1} \geq 8\cdot\frac{a^3-1}{a-1}\cdot\frac{b^3-1}{b-1}\cdot\frac{c^3-1}{c-1}$

Multiply both sides by $(a-1)(b-1)(c-1)$ and recognise the difference of two squares to get:

$(a^3+1)(b^3+1)(c^3+1) \geq 8$ $\implies (abc)^3 + a^3 + b^3 + c^3 + (ab)^3 + (bc)^3 + (ac)^3 + 1 \geq 8$ $\implies a^3 + b^3 + c^3 + (ab)^3 + (bc)^3 + (ac)^3 \geq 6$

Now use AM-GM on each of $a^3 + b^3 + c^3$ and $(ab)^3 + (bc)^3 + (ac)^3$ to obtain:

$a^3 + b^3 + c^3 \geq 3$

$(ab)^3 + (bc)^3 + (ac)^3 \geq 3$

which combines to give the final result.

See Also

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