Difference between revisions of "2009 USAMO Problems/Problem 4"
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Prove that <math>\text{max}(a_1, a_2, ... ,a_n) \le 4 \text{min}(a_1, a_2, ... , a_n)</math>. | Prove that <math>\text{max}(a_1, a_2, ... ,a_n) \le 4 \text{min}(a_1, a_2, ... , a_n)</math>. | ||
− | == Solution == | + | == Solution 1== |
Assume without loss of generality that <math>a_1 \geq a_2 \geq \cdots \geq a_n</math>. Now we seek to prove that <math>a_1 \le 4a_n</math>. | Assume without loss of generality that <math>a_1 \geq a_2 \geq \cdots \geq a_n</math>. Now we seek to prove that <math>a_1 \le 4a_n</math>. | ||
Line 20: | Line 20: | ||
as desired. | as desired. | ||
− | + | == Solution 2== | |
Assume without loss of generality that <math>a_1 \geq a_2 \geq \cdots \geq a_n</math>. | Assume without loss of generality that <math>a_1 \geq a_2 \geq \cdots \geq a_n</math>. | ||
Using the Cauchy–Bunyakovsky–Schwarz inequality and the inequality given, <cmath>\begin{align*} | Using the Cauchy–Bunyakovsky–Schwarz inequality and the inequality given, <cmath>\begin{align*} | ||
− | &\left(a_1+a_2 | + | &\left(a_1+a_2 + ... +a_n + 3a_n -\frac{3a_1}{4}\right)\left({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}\right) \\ |
− | =&\left(\frac{a_1}{4}+a_2 | + | =&\left(\frac{a_1}{4}+a_2 + ... +a_{n-1}+4a_n\right)\left({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}\right) \\ |
\ge& \left(\frac{1}{2}+n-2+2\right)^2 \\ | \ge& \left(\frac{1}{2}+n-2+2\right)^2 \\ | ||
=&\left(n+\frac{1}{2}\right)^2 \\ | =&\left(n+\frac{1}{2}\right)^2 \\ | ||
− | \ge& (a_1+a_2 | + | \ge& (a_1+a_2 + ... +a_{n})\left({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}\right).\end{align*}</cmath> |
+ | (Note that <math>n-2 \ge 0</math> since <math>n \ge 2</math> as given!) | ||
This implies that <math>3a_n -\frac{3a_1}{4} \ge 0 \iff 4a_n \ge a_1</math> as desired. | This implies that <math>3a_n -\frac{3a_1}{4} \ge 0 \iff 4a_n \ge a_1</math> as desired. | ||
+ | |||
+ | ~Deng Tianle, username: Leole | ||
== See Also == | == See Also == |
Latest revision as of 23:50, 13 July 2024
Contents
Problem
For let , , ..., be positive real numbers such that
Prove that .
Solution 1
Assume without loss of generality that . Now we seek to prove that .
By the Cauchy-Schwarz Inequality, Since , clearly , dividing yields:
as desired.
Solution 2
Assume without loss of generality that . Using the Cauchy–Bunyakovsky–Schwarz inequality and the inequality given, (Note that since as given!) This implies that as desired.
~Deng Tianle, username: Leole
See Also
2009 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.