2015 is coming!

Algebra Level 3

If a 1 a 2 + a 3 + . . . + a 2015 + a 2 a 1 + a 3 + . . . + a 2015 + a 3 a 1 + a 2 + a 4 + . . . + a 2015 + . . . + a 2014 a 1 + a 2 + . . . + a 2013 + a 2015 + a 2015 a 1 + a 2 + . . . + a 2014 ≥ K \frac { { a }_{ 1 } }{ { a }_{ 2 }+{ a }_{ 3 }+...+{ a }_{ 2015 } } +\frac { { a }_{ 2 } }{ { a }_{ 1 }+{ a }_{ 3 }+...+{ a }_{ 2015 } } +\frac { { a }_{ 3 } }{ { a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 4 }+...+{ a }_{ 2015 } } +...+\frac { { a }_{ 2014 } }{ { a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ 2013 }+{ a }_{ 2015 } } +\frac { { a }_{ 2015 } }{ { a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ 2014 } } \ge K find K K

Details and assumptions:

  • a 1 , a 2 , . . . , a 2014 , a 2015 { a }_{ 1 },{ a }_{ 2 },...,{ a }_{ 2014 },{ a }_{ 2015 } are positive real numbers.

  • Give your answer to 4 decimals

This problem was inspired by N e s b i t t . . . . . . . n o Nesbitt.......no by Kristian Vasilev


The answer is 1.0005.

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Martin Nikolov by Martin Nikolov , 20, Bulgaria

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4 solutions

Shubhendra Singh
Dec 20, 2014

Let a 1 ≥ a 2 ≥ a 3 ≥ a 4 . . . . . . . . . . . . . . . . ≥ a 2015 a_{1} \geq a_{2} \geq a_{3} \geq a_{4} ................\geq a_{2015}

By this

1 a 2 + a 3 + a 4 + . . . . + a 2015 ≥ 1 a 1 + a 3 + a 4 + . . . + a 2015 ≥ 1 a 1 + a 2 + a 4 . . . . . . + a 2015 . . . ≥ 1 a 1 + a 2 + a 3 . . . + a 2014 \dfrac{1}{a_{2}+a_{3}+a_{4}+ ....+a_{2015}} \geq \dfrac{1}{a_{1}+a_{3}+a_{4}+ ... +a_{2015}} \geq \dfrac{1}{a_{1}+a_{2}+ a_{4} ...... +a_{2015 }} ... \geq \dfrac{1}{a_{1}+a_{2}+a_{3} ...+a_{2014}}

Apply R e − a r r a n g e m e n t i n e q u a l i t y Re-arrangement inequality to get 2014 ( 1 a 2 + a 3 + a 4 + . . . + a 2015 + 1 a 1 + a 3 + a 4 + . . . + a 2015 + 1 a 1 + a 2 + a 4 . . . + a 2015 . . . + 1 a 1 + a 2 + a 3 . . . + a 2014 ) ≥ 2015 ( a 1 + a 2 + a 3 + . . . + a 2015 a 1 + a 2 + a 3 + . . . + a 2015 ) 2014 ( \dfrac{1}{a_{2}+a_{3}+a_{4}+...+a_{2015}}+ \dfrac{1}{a_{1}+a_{3}+a_{4}+...+a_{2015}} + \dfrac{1}{a_{1}+a_{2}+ a_{4}...+a_{2015}}...+\dfrac{1}{a_{1}+a_{2}+a_{3}...+a_{2014}} ) \geq 2015(\dfrac{a_{1}+a_{2}+a_{3}+...+a_{2015}}{a_{1}+a_{2}+a_{3}+...+a_{2015}} )

1 a 2 + a 3 + a 4 + . . . + a 2015 + 1 a 1 + a 3 + a 4 + . . . + a 2015 + 1 a 1 + a 2 + a 4 . . . + a 2015 . . . + 1 a 1 + a 2 + a 3 . . . + a 2014 ≥ 2015 2014 \dfrac{1}{a_{2}+a_{3}+a_{4}+ ...+a_{2015}}+ \dfrac{1}{a_{1}+a_{3}+a_{4}+ ...+a_{2015} }+ \dfrac{1}{a_{1}+a_{2}+ a_{4} ...+a_{2015} }...+\dfrac{1}{a_{1}+a_{2}+a_{3}...+a_{2014}} \geq \dfrac{2015}{2014}

By this K = 2015 2014 K=\dfrac{2015}{2014}

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Generalizing the question, we get ∑ i = 1 n ( a i ( ∑ i = 1 n a i ) − a i ) ≥ n n − 1 \huge\displaystyle\sum_{i=1}^{n} \left( \dfrac{a_i}{\left( \displaystyle\sum_{i=1}^{n} a_i \right) - a_i} \right) \geq\dfrac{n}{n-1}

Aneesh Kundu - 6 years, 5 months ago

Nice!

∑ 1 2015 = x \displaystyle \sum_{1}^{2015} = x

a 1 x − a 1 + a 2 x − a 2 + . . . . + a 2015 x − a 2015 \dfrac{a_{1}}{x - a_{1}} + \dfrac{a_{2}}{x - a_{2}} + .... + \dfrac{a_{2015}}{x - a_{2015}}

x x − a 1 − 1 + x x − a 2 − 1 + . . x x − a 2015 − 1 = ∑ i = 1 2015 ( x x − a i − 1 ) \dfrac{x}{x - a_{1}} - 1 + \dfrac{x}{x - a_{2}} - 1 + .. \dfrac{x}{x - a_{2015}} - 1 = \displaystyle \sum_{i=1}^{2015}( \dfrac{x}{x - a_{i}} - 1)

∑ i = 1 2015 x x − a i 2015 ≥ 2015 ∑ i = 1 2015 x − a i x \displaystyle \dfrac{\sum_{i=1}^{2015} \dfrac{x}{x - a_{i}}}{2015} \geq \dfrac{2015}{ \displaystyle \sum_{i=1}^{2015} \dfrac{x - a_{i}}{x}}

∑ i = 1 2015 x x − a i 2015 ≥ 2015 x 2014 x \displaystyle \dfrac{\sum_{i=1}^{2015} \dfrac{x}{x - a_{i}}}{2015} \geq \dfrac{2015x}{2014x}

∑ i = 1 2015 ( x x − a i − 1 ) ≥ 2015 2014 \displaystyle \sum_{i=1}^{2015} (\dfrac{x}{x - a_{i}} -1) \geq \dfrac{2015}{2014}

U Z - 6 years, 5 months ago
Ap V
Aug 1, 2015

Min will occur when all the ai's become equal . so it can be an objective trick

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Priyesh Pandey
Jan 3, 2015

since question stated that each term should be a positive real no. , hence i put each term equals to 1.. since in this inequality equality will hold when each term will be equal .. thus by doing this i got the result 2015/2014

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Mattapalli Ram
Dec 22, 2014

if we exchange any a i a_{i} with any a j a_{j} , the sum value will be same. So the sum is symmetric and the minimum occurs when all a i a_{i} 's are same. which gives 2015/2014.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

It isn't necessary that minimum in a symmetric function/sum will always occur when all the variables are same. Check out this discussion on the topic.

Prasun Biswas - 6 years, 3 months ago

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