Difference between revisions of "1993 USAMO Problems/Problem 5"
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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> | ||
+ | |||
+ | ==Solution== | ||
+ | Notice that because | ||
+ | <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> | ||
+ | <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> | ||
+ | <cmath> = a_0 a_n + (a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n),</cmath> | ||
+ | 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 | ||
+ | <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> | ||
+ | or | ||
+ | <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> | ||
+ | |||
+ | Fortunately, this is an easy inequality. Indeed, from AM-GM applied on each group of terms we have | ||
+ | <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> | ||
+ | and so it suffices to prove | ||
+ | <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> | ||
+ | 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 | ||
+ | <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> | ||
+ | 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 | ||
+ | <cmath>a_0 a_2 \le a_1^2</cmath> | ||
+ | <cmath>a_1 a_3 \le a_2^2</cmath> | ||
+ | <cmath>a_2 a_4 \le a_3^2</cmath> | ||
+ | <cmath>\dots</cmath> | ||
+ | <cmath>a_{n-3} a_{n-1} \le a_{n-2}^2</cmath> | ||
+ | <cmath>a_{n-2} a_n \le a_{n-1}^2.</cmath> | ||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | --User suli, April 15, 2015. | ||
+ | |||
+ | == See Also == | ||
+ | {{USAMO box|year=1993|num-b=4|after=Last Problem}} | ||
+ | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356413#p356413 Discussion on AoPS/MathLinks] | ||
+ | |||
+ | [[Category:Olympiad Algebra Problems]] | ||
+ | [[Category:Olympiad Inequality Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 18:08, 1 June 2020
Problem 5
Let be a sequence of positive real numbers satisfying for . (Such a sequence is said to be log concave.) Show that for each ,
Solution
Notice that because we may subtract from both sides of the inequality and observe that it is sufficient to prove that or
Fortunately, this is an easy inequality. Indeed, from AM-GM applied on each group of terms we have and so it suffices to prove or, after taking both sides to the power, simplifying, and taking the -th root of both sides, to prove This easily follows from the Fact that for . Indeed, we are given that Multiply all inequalities together and cancel to give . Similarly, by multiplying all inequalities except the first and the last, we deduce that , and a simple induction argument proves the verity of the Fact for , and so by the Commutative Property the Fact is true for all , as desired. Now multiply each inequality of the Fact for to give the desired result.
Note: The Fact can be generalized into a Lemma: whenever and . 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 (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.