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Algebra Level 5

There are 2011 positive numbers with both their sum and the sum of their reciprocals equal to 2012. Let x x be one of these numbers, if the maximum value of x + 1 x x +\dfrac{1}{x} can be expressed as a b \frac ab for coprime positive integers, submit your answer as a b a-b .


The answer is 6033.

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1 solution

Abhijeet Verma
Sep 7, 2015

L e t p 1 , p 2 . . . . p 2010 b e t h e 2010 n u m b e r s d i s t i n c t f r o m x . T h e n , p 1 + p 2 + . . . . . . + p 2010 = 2012 x a n d 1 p 1 + 1 p 2 + . . . . + 1 p 2010 = 2012 1 x A p p l y i n g C a u c h y S c h w a r z / A M G M I n e q u a l i t y , Let\quad { p }_{ 1 }{ ,p }_{ 2 }....{ p }_{ 2010 }\quad be\quad the\quad 2010\quad numbers\quad distinct\quad from\quad x.\\ Then,\quad { p }_{ 1 }+{ p }_{ 2 }+{ ......+p }_{ 2010 }\quad =\quad 2012-x\\ and\quad \frac { 1 }{ { p }_{ 1 } } +\frac { 1 }{ { p }_{ 2 } } +....+\frac { 1 }{ { p }_{ 2010 } } \quad =\quad 2012-\frac { 1 }{ x } \\ Applying\quad Cauchy-Schwarz\quad /\quad AM-GM\quad Inequality, ( i = 1 2010 p i ) ( i = 1 2010 1 p i ) = ( 2012 x ) ( 2012 1 x ) 2010 2 2012 2 2012 ( x + 1 x ) + 1 2010 2 0 ( x + 1 x ) 8045 2012 \left( \sum _{ i=1 }^{ 2010 }{ { p }_{ i } } \right) \left( \sum _{ i=1 }^{ 2010 }{ \frac { 1 }{ { p }_{ i } } } \right) =(2012-x)(2012-\frac { 1 }{ x } )\quad \ge \quad { 2010 }^{ 2 }\\ \Longrightarrow \quad { 2012 }^{ 2 }-\quad 2012(x+\frac { 1 }{ x } )+1-{ 2010 }^{ 2 }\quad \ge \quad 0\\ \\ \Longrightarrow \quad (x+\frac { 1 }{ x } )\le \frac { 8045 }{ 2012 } T h u s , 8045 2012 = 6033 Thus,\quad 8045-2012=\boxed{6033}

The answer is a maximum of 4.

adding the two equalities, we get [sum{i=1,2010} (pi + 1/pi) ] + (x + 1/x) = 4024

Thus

maximize (x + 1/x) is equivalent to minimize [sum{i=1,2010} (pi + 1/pi) ]

since p is positive, min value of p + 1/p = 2 (when p=1)

==> min [sum{i=1,2010} (pi + 1/pi) ] = 4020

==> max (x+1/x) = 4

your solution neglects the fact that the variables p1 to p2010 are common in both the inequalities. Your solution would have been correct if the two series were

p1 + p2 + .. + p2010 + x = 2012

&

1/q1 + 1/q2 + ...+1/q2010 + 1/x = 2012

prasun bansal - 5 years, 8 months ago

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@prasun bansal You have taken p + 1/p =2 ,which implies p= 1/p =1. Now, the first equation yields x=2 , while the second equation yields x=0.5.

Abhijeet Verma - 5 years, 3 months ago

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