# Difference between revisions of "2011 AIME II Problems/Problem 9"

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− | Note that | + | 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> |

+ | |||

+ | An example is: | ||

+ | <cmath> | ||

+ | \begin{align*} | ||

+ | x_3 &= x_6 = \frac16 \\ | ||

+ | x_1 &= x_2 = \frac{15 - \sqrt{220}}{30} \\ | ||

+ | x_5 &= x_4 = \frac{15 + \sqrt{220}}{30} | ||

+ | \end{align*} | ||

+ | </cmath> |

## Revision as of 23:38, 1 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 constraint expression is equal to with —for example, by choosing and small enough—so our answer is

An example is: