Sounds "related" to Ineq #1

Algebra Level 3

Let a , b , c a, b, c be positive reals so that a + b + c = 69 a + b + c = 69 and a b c = 12167 abc = 12167 . Find 2015 a + 2016 b + 2017 c 2015a + 2016b + 2017c .


The answer is 139104.

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Steven Jim by Steven Jim , 17, Vietnam

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1 solution

It should be mentioned that a , b , c > 0 a,b,c>0 , By noting that ( a + b + c ) 3 ≥ 27 a b c (a+b+c)^3 \ge 27abc holds in this case that is ( a + b + c ) 3 = 27 a b c (a+b+c)^3=27abc we conclude that a = b = c a=b=c by equality conditions of AM-GM . Thus the answer is 23 ( 2015 + 2016 + 2017 ) = 139104 23(2015+2016+2017)=\boxed{139104}

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Thanks! I've fixed the problem. By the way, good solution!

Another way of solving this is to use AM - GM: 69 = a + b + c ≥ 3 a b c 3 = > a b c 3 ≤ 23 = > a b c ≤ 2 3 3 = 12167 69=a+b+c\ge 3\sqrt [ 3 ]{ abc } => \sqrt [ 3 ]{ abc } \le 23 => abc \le 23^{3} = 12167 .

Steven Jim - 4 years, 1 month ago

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