A question on finding maximum
IF a+b+c=1,then find maximum value of ab+bc+ca? (a,b,c are positive numbers)
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Shubham Bagrecha
8 years, 2 months ago
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2 votes
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(a+b+c)2=1 a2+b2+c2+2ab+2bc+2ca=1 a2+b2+c2−ab−bc−ca+3ab+3bc+3ca=1 21[(a−b)2+(b−c)2+(c−a)2]+3(ab+bc+ca)=1 3(ab+bc+ca) will be maximum when 21[(a−b)2+(b−c)2+(c−a)2]=0 i.e. is minimum =0. Therefore max. ab+bc+ca=31
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Nice proof!
solution?
square a+b+c to get a^2 + b^2 + c^2 + 2(ab+bc+ca) = 1. Note that a^2 + b^2 + c^2 >= ab + bc + ca by AM-GM. thus ab + bc + ca <= 1/3 with equality iff a = b =c = 1/3
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The result also follows from the Rearrangement-Inequality
You can also use LaGrange Multipliers to solve for the maximum value right?
AM-GM kills it
1/3 is the maximum
0.33
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0.333333... To be precise (or simply 1/3, you meant 33/100)
1/3
The max. value of ab + bc + ca is 0.
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that's wrong
maximum is never 0
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Not always true
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E.g. Maximum of -x^2 where x∈R is obviously 0, attained when x = 0.
i know, bt m talking about this case