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

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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>
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<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==

Revision as of 14:07, 29 January 2022

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

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