Difference between revisions of "2011 USAMO Problems/Problem 1"
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<cmath>\frac{ab + 1}{(a + b)^2} + \frac{bc + 1}{(b + c)^2} + \frac{ca + 1}{(c + a)^2} \ge 3.</cmath> | <cmath>\frac{ab + 1}{(a + b)^2} + \frac{bc + 1}{(b + c)^2} + \frac{ca + 1}{(c + a)^2} \ge 3.</cmath> | ||
− | ==Solution== | + | ==Solutions== |
+ | ===Solution 1=== | ||
Since | Since | ||
<cmath> | <cmath> | ||
Line 57: | Line 58: | ||
We note that the product of the three (positive) terms is 1/8, therefore by AM-GM their mean is at least 1/2, and thus their sum is at least 3/2 and we are done. | We note that the product of the three (positive) terms is 1/8, therefore by AM-GM their mean is at least 1/2, and thus their sum is at least 3/2 and we are done. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | Rearranging the condition yields that | ||
+ | <cmath>a^2 + b^2 + c^2 +ab+bc+ac \le 2</cmath> | ||
+ | |||
+ | Now note that | ||
+ | <cmath>\frac{2ab+2}{(a+b)^2} \ge \frac{2ab+a^2 + b^2 + c^2 +ab+bc+ac}{(a+b)^2}=\frac{(a+b)^2 + (c+a)(c+b)}{(a+b)^2}</cmath> | ||
+ | |||
+ | Summing this for all pairs of <math>\{ a,b,c \}</math> gives that | ||
+ | <cmath>\sum_{cyc} \frac{2ab+2}{(a+b)^2} \ge 3+ \sum_{cyc}\frac{(c+a)(c+b)}{(a+b)^2} \ge 6</cmath> | ||
+ | |||
+ | By AM-GM. Dividing by <math>2</math> gives the desired inequality. | ||
+ | |||
+ | {{MAA Notice}} | ||
+ | |||
+ | ==See also== | ||
+ | {{USAMO newbox|year=2011|beforetext=|before=First Problem|num-a=2}} | ||
+ | {{USAJMO newbox|year=2011|num-b=1|num-a=3}} | ||
+ | [[Category:Olympiad Algebra Problems]] | ||
+ | [[Category:Olympiad Inequality Problems]] |
Latest revision as of 13:57, 24 October 2020
Problem
Let , , be positive real numbers such that . Prove that
Solutions
Solution 1
Since it is natural to consider a change of variables: with the inverse mapping given by: With this change of variables, the constraint becomes while the left side of the inequality we need to prove is now
Therefore it remains to prove that
We note that the product of the three (positive) terms is 1/8, therefore by AM-GM their mean is at least 1/2, and thus their sum is at least 3/2 and we are done.
Solution 2
Rearranging the condition yields that
Now note that
Summing this for all pairs of gives that
By AM-GM. Dividing by gives the desired inequality.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2011 USAMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
2011 USAJMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |