Difference between revisions of "2011 AIME II Problems/Problem 9"
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| − | Note that neither the constraint nor the expression we need to maximize involves products <math>x_i x_j</math> with <math>i - j \equiv 3 \pmod 6</math>. Factoring out say <math>x_1</math> and <math>x_4</math> we see that the constraint is <math>x_1(x_3x_5) + x_4(x_2x_6) \ge {\scriptstyle\frac1{540}}</math>, while the expression we want to maximize is <math>x_1(x_2x_3 + x_5x_6 + x_6x_2) + x_4(x_2x_3 + x_5x_6 + x_3x_5)</math>. Adding the left side of the constraint to the expression we get: <math>(x_1 + x_4)(x_2x_3 + x_5x_6 + x_6x_2 + x_3x_5) = (x_1 + x_4)(x_2 + x_5)(x_3 + x_6)</math>. This new expression is the product of three non-negative terms whose sum is equal to 1. By AM-GM this product is at most <math>\scriptstyle\frac1{27}</math>. Since we have added at least <math>\scriptstyle\frac1{540}</math> the desired maximum is at most <math>\scriptstyle\frac1{27} - \frac1{540} = \frac{19}{540}</math>. It is easy to see that this upper bound can in fact be achieved by ensuring that constraint expression is equal to <math>\scriptstyle\frac1{540}</math> with <math>x_1 + x_4 = x_2 + x_5 = x_3 + x_6 = \scriptstyle\frac13</math>—for example, by choosing <math>x_1</math> and <math>x_2</math> small enough—so our answer is <math>540 + 19 = \fbox{559}.</math> | + | Note that neither the constraint nor the expression we need to maximize involves products <math>x_i x_j</math> with <math>i - j \equiv 3 \pmod 6</math>. Factoring out say <math>x_1</math> and <math>x_4</math> we see that the constraint is <math>x_1(x_3x_5) + x_4(x_2x_6) \ge {\scriptstyle\frac1{540}}</math>, while the expression we want to maximize is <math>x_1(x_2x_3 + x_5x_6 + x_6x_2) + x_4(x_2x_3 + x_5x_6 + x_3x_5)</math>. Adding the left side of the constraint to the expression we get: <math>(x_1 + x_4)(x_2x_3 + x_5x_6 + x_6x_2 + x_3x_5) = (x_1 + x_4)(x_2 + x_5)(x_3 + x_6)</math>. This new expression is the product of three non-negative terms whose sum is equal to 1. By AM-GM this product is at most <math>\scriptstyle\frac1{27}</math>. Since we have added at least <math>\scriptstyle\frac1{540}</math> the desired maximum is at most <math>\scriptstyle\frac1{27} - \frac1{540} = \frac{19}{540}</math>. It is easy to see that this upper bound can in fact be achieved by ensuring that the constraint expression is equal to <math>\scriptstyle\frac1{540}</math> with <math>x_1 + x_4 = x_2 + x_5 = x_3 + x_6 = \scriptstyle\frac13</math>—for example, by choosing <math>x_1</math> and <math>x_2</math> small enough—so our answer is <math>540 + 19 = \fbox{559}.</math> |
An example is: | An example is: | ||
Revision as of 14:42, 2 April 2011
Problem 9
Let 
be non-negative real numbers such that 
, and 
. Let 
and 
be positive relatively prime integers such that 
is the maximum possible value of
. Find 
.
Solution
Note that neither the constraint nor the expression we need to maximize involves products 
with 
. Factoring out say 
and 
we see that the constraint is 
, while the expression we want to maximize is 
. Adding the left side of the constraint to the expression we get: 
. This new expression is the product of three non-negative terms whose sum is equal to 1. By AM-GM this product is at most 
. Since we have added at least 
the desired maximum is at most 
. It is easy to see that this upper bound can in fact be achieved by ensuring that the constraint expression is equal to with 
—for example, by choosing and 
small enough—so our answer is 
An example is:
