If a 1 , a 2 , a 3 . . . . . . . a 2 0 1 5 > 0 , a 1 + a 2 + . . . + a 2 0 1 5 = 1
and the minimum value of
a 1 1 + a 2 1 + a 3 1 + . . . + a 2 0 1 5 1 = x
what is the value of x
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we can used Jenson's inequality
Let y = f ( x ) = x 1 x ∈ ( 0 , 1 ) .
Consider 2015 point's on the curve So after joining them an polygon is formed whose Centroid is G and which lies inside the polygon , and a point P which lies on the curve with same x-coordinate as that of centroid So
y G ≥ y p n ∑ f ( x i ) ≥ f ( n ∑ x i ) ∑ f ( x i ) ≥ n f ( n 1 ) ( p u t n = 2 0 1 5 ) ∑ f ( x i ) ≥ 2 0 1 5 2 k = 2 0 1 5 .
How did you include the graph in your solution? :o
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hi Aritra , First Upload your image on the sites Like : imgur,com
and then Copy the image URL ( By right click on image )
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! [Type Title] (Post image URL here)
Don't give Space between two brackets and exclamation ( ! ) mark
It can also be done by Titu's Lemma
I did this through a simple application of Titu's Lemma. The key is to treat all the numerators as if they were the square of 1, rather than just 1. Easiest solution, in my opinion.
@Ryan Tamburrino can you please write easiest solution in your mind style.
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L e t u s u s e A . M − H . M : 2 0 1 5 a 1 1 + a 2 1 + a 3 1 + . . . a 2 0 1 5 1 ≥ a 1 + a 2 + a 3 + . . . . a 2 0 1 5 2 0 1 5 B u t , a 1 + a 2 + a 3 + . . . . a 2 0 1 5 = 1 ⟹ 2 0 1 5 a 1 1 + a 2 1 + a 3 1 + . . . a 2 0 1 5 1 ≥ 2 0 1 5 ⟹ a 1 1 + a 2 1 + a 3 1 + . . . a 2 0 1 5 1 ≥ 2 0 1 5 2 ⟹ a n s w e r = 2 0 1 5 2 = 2 0 1 5 . A . M − H . M w o r k s t h i s w a y : n ( a 1 + a 2 + . . . a n ) ≥ a 1 1 + a 2 1 + . . . a n 1 n