Difference between revisions of "1993 USAMO Problems/Problem 5"

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== Problem 4==
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== Problem 5==
  
 
Let <math>a_0, a_1, a_2,\cdots</math> be a sequence of positive real numbers satisfying <math>a_{i-1}a_{i+1}\le a^2_i</math>
 
Let <math>a_0, a_1, a_2,\cdots</math> be a sequence of positive real numbers satisfying <math>a_{i-1}a_{i+1}\le a^2_i</math>
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<center><math>\frac{a_0+\cdots+a_n}{n+1}\cdot\frac{a_1+\cdots+a_{n-1}}{n-1}\ge\frac{a_0+\cdots+a_{n-1}}{n}\cdot\frac{a_1+\cdots+a_{n}}{n}</math>.</center>
 
<center><math>\frac{a_0+\cdots+a_n}{n+1}\cdot\frac{a_1+\cdots+a_{n-1}}{n-1}\ge\frac{a_0+\cdots+a_{n-1}}{n}\cdot\frac{a_1+\cdots+a_{n}}{n}</math>.</center>
  
== Resources ==
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==Solution==
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Notice that because
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<cmath>(a_0 + a_1 + \dots + a_{n-1})(a_1 + a_2 + \dots + a_n) = (a_0 + (a_1 + \dots + a_{n-1})(-a_0 + (a_0 + a_1 + \dots + a_n))</cmath>
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<cmath> = -a_0^2 + a_0 ((a_0 + a_1 + \dots + a_n) - (a_1 + a_2 + \dots + a_{n-1})) + (a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n)</cmath>
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<cmath> = a_0 a_n + (a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n),</cmath>
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we may subtract <math>\dfrac{(a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n)}{n^2}</math> from both sides of the inequality and observe that it is sufficient to prove that
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<cmath>\frac{(a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n)}{(n^2 - 1)n^2} \ge \frac{a_0 a_n}{n^2},</cmath>
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or
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<cmath>(a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n) \ge (n^2 - 1) a_0 a_n.</cmath>
  
{{USAMO box|year=1993|num-b=3|num-a=5}}
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Fortunately, this is an easy inequality. Indeed, from AM-GM applied on each group of terms we have
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<cmath>(a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n) \ge (n^2 - 1) \sqrt[n-1]{a_1 a_2 \dots a_{n-1}} \sqrt[n+1]{a_0 a_1 \dots a_n},</cmath>
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and so it suffices to prove
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<cmath>\sqrt[n-1]{a_1 a_2 \dots a_{n-1}} \sqrt[n+1]{a_0 a_1 \dots a_n} \ge a_0 a_n,</cmath>
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or, after taking both sides to the <math>(n^2 - 1)</math> power, simplifying, and taking the <math>n</math>-th root of both sides, to prove
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<cmath>a_1^2 a_2^2 a_3^2 \dots a_{n-1}^2 \ge a_0^{n-1} a_n^{n-1}.</cmath>
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This easily follows from the Fact that <math>a_0 a_n \le a_i a_{n-i}</math> for <math>1 \le i \le n-1</math>. Indeed, we are given that
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<cmath>a_0 a_2 \le a_1^2</cmath>
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<cmath>a_1 a_3 \le a_2^2</cmath>
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<cmath>a_2 a_4 \le a_3^2</cmath>
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<cmath>\dots</cmath>
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<cmath>a_{n-3} a_{n-1} \le a_{n-2}^2</cmath>
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<cmath>a_{n-2} a_n \le a_{n-1}^2.</cmath>
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Multiply all inequalities together and cancel <math>a_1, a_2^2, a_3^2, \dots, a_{n-2}^2, a_{n-1}</math> to give <math>a_0 a_n \le a_1 a_{n-1}</math>. Similarly, by multiplying all inequalities except the first and the last, we deduce that <math>a_2 a_{n-2} \le a_1 a_{n-1} \ge a_0 a_n</math>, and a simple induction argument proves the verity of the Fact for <math>1 \le i \le \frac{n}{2}</math>, and so by the Commutative Property the Fact is true for all <math>1 \le i \le n-1</math>, as desired. Now multiply each inequality of the Fact for <math>i = 1, 2, 3, \dots, n-1</math> to give the desired result.
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Note: The Fact can be generalized into a Lemma: <math>a_x a_w \le a_y a_z</math> whenever <math>x \le y \le z \le w</math> and <math>x + w = y + z</math>. The proof is similar to that of the Fact and is left as an exercise to the reader.
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--User suli, April 15, 2015.
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== See Also ==
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{{USAMO box|year=1993|num-b=4|after=Last Problem}}
 
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356413#p356413 Discussion on AoPS/MathLinks]
 
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356413#p356413 Discussion on AoPS/MathLinks]
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[[Category:Olympiad Algebra Problems]]
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[[Category:Olympiad Inequality Problems]]
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{{MAA Notice}}

Latest revision as of 18:08, 1 June 2020

Problem 5

Let $a_0, a_1, a_2,\cdots$ be a sequence of positive real numbers satisfying $a_{i-1}a_{i+1}\le a^2_i$ for $i = 1, 2, 3,\cdots$ . (Such a sequence is said to be log concave.) Show that for each $n > 1$,

$\frac{a_0+\cdots+a_n}{n+1}\cdot\frac{a_1+\cdots+a_{n-1}}{n-1}\ge\frac{a_0+\cdots+a_{n-1}}{n}\cdot\frac{a_1+\cdots+a_{n}}{n}$.

Solution

Notice that because \[(a_0 + a_1 + \dots + a_{n-1})(a_1 + a_2 + \dots + a_n) = (a_0 + (a_1 + \dots + a_{n-1})(-a_0 + (a_0 + a_1 + \dots + a_n))\] \[= -a_0^2 + a_0 ((a_0 + a_1 + \dots + a_n) - (a_1 + a_2 + \dots + a_{n-1})) + (a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n)\] \[= a_0 a_n + (a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n),\] we may subtract $\dfrac{(a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n)}{n^2}$ from both sides of the inequality and observe that it is sufficient to prove that \[\frac{(a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n)}{(n^2 - 1)n^2} \ge \frac{a_0 a_n}{n^2},\] or \[(a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n) \ge (n^2 - 1) a_0 a_n.\]

Fortunately, this is an easy inequality. Indeed, from AM-GM applied on each group of terms we have \[(a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n) \ge (n^2 - 1) \sqrt[n-1]{a_1 a_2 \dots a_{n-1}} \sqrt[n+1]{a_0 a_1 \dots a_n},\] and so it suffices to prove \[\sqrt[n-1]{a_1 a_2 \dots a_{n-1}} \sqrt[n+1]{a_0 a_1 \dots a_n} \ge a_0 a_n,\] or, after taking both sides to the $(n^2 - 1)$ power, simplifying, and taking the $n$-th root of both sides, to prove \[a_1^2 a_2^2 a_3^2 \dots a_{n-1}^2 \ge a_0^{n-1} a_n^{n-1}.\] This easily follows from the Fact that $a_0 a_n \le a_i a_{n-i}$ for $1 \le i \le n-1$. Indeed, we are given that \[a_0 a_2 \le a_1^2\] \[a_1 a_3 \le a_2^2\] \[a_2 a_4 \le a_3^2\] \[\dots\] \[a_{n-3} a_{n-1} \le a_{n-2}^2\] \[a_{n-2} a_n \le a_{n-1}^2.\] Multiply all inequalities together and cancel $a_1, a_2^2, a_3^2, \dots, a_{n-2}^2, a_{n-1}$ to give $a_0 a_n \le a_1 a_{n-1}$. Similarly, by multiplying all inequalities except the first and the last, we deduce that $a_2 a_{n-2} \le a_1 a_{n-1} \ge a_0 a_n$, and a simple induction argument proves the verity of the Fact for $1 \le i \le \frac{n}{2}$, and so by the Commutative Property the Fact is true for all $1 \le i \le n-1$, as desired. Now multiply each inequality of the Fact for $i = 1, 2, 3, \dots, n-1$ to give the desired result.

Note: The Fact can be generalized into a Lemma: $a_x a_w \le a_y a_z$ whenever $x \le y \le z \le w$ and $x + w = y + z$. The proof is similar to that of the Fact and is left as an exercise to the reader.

--User suli, April 15, 2015.

See Also

1993 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Last Problem
1 2 3 4 5
All USAMO Problems and Solutions

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