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

Latest revision as of 16:08, 13 August 2023

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 $a_2 a_{n-2} \ge a_1 a_{n-1} \ge a_0 a_n$ , 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.

@@ Line 5: / Line 5: @@
 each <math>n > 1</math>,
-<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>
+<cmath>\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}.</cmath>
 ==Solution==
-{{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} \ge 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 ==

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

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

Latest revision as of 16:08, 13 August 2023

Problem 5

Solution

See Also