But it's 2017!

Level 2

Let x 1 , x 2 . . . . , x 2014 x_{1},x_{2}....,x_{2014} be positive real numbers such that

∑ j = 1 2014 x j = 1 \large \sum _{ j=1 }^{ 2014 }{ { x }_{ j } } =1 .

Then,

x 1 2 1 − x 1 + x 2 2 1 − x 2 . . . . . x 2014 2 1 − x 2014 ≥ 1 K \large \frac { { { x }_{ 1 } }^{ 2 } }{ 1-{ x }_{ 1 } } +\frac { { { x }_{ 2 } }^{ 2 } }{ 1-{ x }_{ 2 } } .....\frac { { { { x }_{ 2014 } }^{ 2 } } }{ 1-{ x }_{ 2014 } } \ge \frac { 1 }{ K } .

What is the smallest value of K \large K ?


The answer is 2013.

Achal Jain by Achal Jain , 19, India

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1 solution

Md Zuhair
Nov 22, 2017

By Titu's Lemma,

x 1 2 1 − x 1 + x 2 2 1 − x 2 . . . . . x 2014 2 1 − x 2014 ≥ ( x 1 + x 2 + . . . x 2014 ) 2 1 − x 1 + 1 − x 2 + . . . + 1 − x 2014 = ( x 1 + x 2 + . . . x 2014 ) 2 2014 − x 1 − x 2 − . . . x 2014 = 1 2014 − 1 = 1 2013 \large{ \frac { { { x }_{ 1 } }^{ 2 } }{ 1-{ x }_{ 1 } } +\frac { { { x }_{ 2 } }^{ 2 } }{ 1-{ x }_{ 2 } } .....\frac { { { { x }_{ 2014 } }^{ 2 } } }{ 1-{ x }_{ 2014 } } \geq \frac{(x_{1}+x_{2}+...x_{2014})^{2}}{1-x_{1}+1-x_{2}+...+1-x_{2014}} = \frac{(x_{1}+x_{2}+...x_{2014})^{2}}{2014-x_{1}-x_{2}-...x_{2014}} = \dfrac{1}{2014-1} = \dfrac{1}{2013}}

So the least value of K = 2013 \boxed{K =2013}

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