Let a 1 , a 2 , a 3 , … , a 1 7 3 be positive real numbers such that a 1 + a 2 + a 3 + … + a 1 7 3 = 1 7 4 . Find the minimum possible value of
a 1 + a 2 a 1 2 + a 2 + a 3 a 2 2 + a 3 + a 4 a 3 2 + … + a 1 7 3 + a 1 a 1 7 3 2 .
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I used Cauchy then I got 8 7
By Cauchy-Schwarz in Engel Form, we have
a 1 + a 2 a 1 2 + a 2 + a 3 a 2 2 + ⋯ + a 1 7 3 + a 1 a 1 7 3 2 ≥ 2 ( a 1 + a 2 + ⋯ + a 1 7 3 ) ( a 1 + a 2 + ⋯ + a 1 7 3 ) 2 = 8 7
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I used jensens and put a 4 in the numerator cuz I accidentally squared the 2 up top. Careless
Yeah. CS is the easiest way to do it :D.
we can also apply AM-HM inequality.
Consider all terms equal for extreme limits
there are 173 positive real numbers and their sum is 174.So from here we can get the minimum value of 1st 172 numbers as 1 and the minimum value of last number as 2.. By doing so the minimum value of the given equation will be 87.33. Since they are asking for the minimum integer number so the answer is 87
Absolutely Wrong Solution.
Abbbssssoooolllluuuttttteeeellly wwwwwwwwwwwwrrrrrrrrrrrrrrrrrrrrrooooooooooooooooonnnnnnnnnnnnnngggggggggg
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Titu's Lemma.