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> | ||
| Line 4: | Line 5: | ||
each <math>n > 1</math>, | each <math>n > 1</math>, | ||
| − | < | + | <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== | ||
| + | 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 == | ||
| + | {{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 17:08, 13 August 2023
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 
,
![\[\frac{a_0+\cdots+a_n}{n+1}\cdot\frac{a_1+\cdots+a_{n-1}}{n-1}\ge\frac{a_0+\cdots+a_{n-1}}{n}\cdot\frac{a_1+\cdots+a_{n}}{n}.\]](http://latex.artofproblemsolving.com/9/e/5/9e5d982a0cde57394fa814a77af453740ec5720a.png)
Solution
Notice that because
![\[(a_0 + a_1 + \dots + a_{n-1})(a_1 + a_2 + \dots + a_n) = (a_0 + (a_1 + \dots + a_{n-1})(-a_0 + (a_0 + a_1 + \dots + a_n))\]](http://latex.artofproblemsolving.com/8/5/a/85a8d3c19527cb6e8936d86424a53cee9c21a461.png)
![\[= -a_0^2 + a_0 ((a_0 + a_1 + \dots + a_n) - (a_1 + a_2 + \dots + a_{n-1})) + (a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n)\]](http://latex.artofproblemsolving.com/b/9/d/b9d7bf65a22adc7db28b02a751401647e300f639.png)
![\[= a_0 a_n + (a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n),\]](http://latex.artofproblemsolving.com/1/7/3/173ad8e7406b58fe1a2dec540af2ea6029dff323.png)
we may subtract 
from both sides of the inequality and observe that it is sufficient to prove that
![\[\frac{(a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n)}{(n^2 - 1)n^2} \ge \frac{a_0 a_n}{n^2},\]](http://latex.artofproblemsolving.com/a/f/5/af55f114515f155d0feb4d4a69372808c3489cfc.png)
or
![\[(a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n) \ge (n^2 - 1) a_0 a_n.\]](http://latex.artofproblemsolving.com/a/3/3/a33412c42d0a1e053fb899fc265598c7188d4924.png)
Fortunately, this is an easy inequality. Indeed, from AM-GM applied on each group of terms we have
![\[(a_1 + a_2 + \dots + a_{n-1})(a_0 + a_1 + \dots + a_n) \ge (n^2 - 1) \sqrt[n-1]{a_1 a_2 \dots a_{n-1}} \sqrt[n+1]{a_0 a_1 \dots a_n},\]](http://latex.artofproblemsolving.com/a/c/0/ac050fa06115f37b2ebaeb29ed222530ad4d8de5.png)
and so it suffices to prove
![\[\sqrt[n-1]{a_1 a_2 \dots a_{n-1}} \sqrt[n+1]{a_0 a_1 \dots a_n} \ge a_0 a_n,\]](http://latex.artofproblemsolving.com/3/c/6/3c65bcbc0fef68bd9c13060c1551bf17a305b7de.png)
or, after taking both sides to the 
power, simplifying, and taking the 
-th root of both sides, to prove
![\[a_1^2 a_2^2 a_3^2 \dots a_{n-1}^2 \ge a_0^{n-1} a_n^{n-1}.\]](http://latex.artofproblemsolving.com/f/7/5/f75589d3fcbbae0376e786e504146b9086e029da.png)
This easily follows from the Fact that 
for 
. Indeed, we are given that
![\[a_0 a_2 \le a_1^2\]](http://latex.artofproblemsolving.com/e/4/f/e4f05636cd929df959ca1cd0933877ced94e934b.png)
![\[a_1 a_3 \le a_2^2\]](http://latex.artofproblemsolving.com/b/d/6/bd6426bf356e86e5ec6e107720f92febad3f3abb.png)
![\[a_2 a_4 \le a_3^2\]](http://latex.artofproblemsolving.com/a/d/7/ad710a42eb806c75df33c6ae12ded39ac8e21936.png)
![\[\dots\]](http://latex.artofproblemsolving.com/1/6/7/16762e47f22f53e2431935f211fa82570b46b6cb.png)
![\[a_{n-3} a_{n-1} \le a_{n-2}^2\]](http://latex.artofproblemsolving.com/8/e/6/8e6bcd3375315b852b7d0aa747e315df7199a666.png)
![\[a_{n-2} a_n \le a_{n-1}^2.\]](http://latex.artofproblemsolving.com/1/8/e/18edef9467650c996ae36bbef5da27876789f01c.png)
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. 
