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A question on finding maximum

IF a+b+c=1,then find maximum value of ab+bc+ca? (a,b,c are positive numbers)

Note by Shubham Bagrecha
4 years, 3 months ago

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\[(a+b+c)^{2}=1\] \[a^{2}+b^{2}+c^{2}+2ab+2bc+2ca=1\] \[a^{2}+b^{2}+c^{2}-ab-bc-ca+3ab+3bc+3ca=1\] \[\frac {1}{2}[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]+3(ab+bc+ca)=1\] \[3(ab+bc+ca)\] will be maximum when \[\frac {1}{2}[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]=0\] i.e. is minimum =0. Therefore max. \[ab+bc+ca=\frac {1}{3}\] Shubham Srivastava · 4 years, 3 months ago

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@Shubham Srivastava Nice proof! Rohan Rao · 4 years, 3 months ago

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AM-GM kills it Taufiq Hakim Rac'madhan · 4 years, 3 months ago

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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 Gabriel Wong · 4 years, 3 months ago

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@Gabriel Wong The result also follows from the Rearrangement-Inequality Adhiraj Mandal · 4 years, 3 months ago

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solution? Shubham Bagrecha · 4 years, 3 months ago

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You can also use LaGrange Multipliers to solve for the maximum value right? Eric Wu · 4 years, 3 months ago

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1/3 Abhimanyu Singla · 4 years, 3 months ago

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1/3 is the maximum Ed Mañalac · 4 years, 3 months ago

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0.33 Nishit Goyal · 4 years, 3 months ago

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@Nishit Goyal 0.333333... To be precise (or simply 1/3, you meant 33/100) Zi Song Yeoh · 4 years, 3 months ago

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The max. value of ab + bc + ca is 0. Sarthak Nandan · 4 years, 3 months ago

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@Sarthak Nandan that's wrong Abhimanyu Singla · 4 years, 3 months ago

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@Sarthak Nandan maximum is never 0 Abhimanyu Singla · 4 years, 3 months ago

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@Abhimanyu Singla Not always true Zi Song Yeoh · 4 years, 3 months ago

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@Zi Song Yeoh E.g. Maximum of -x^2 where \(x \in \mathbb{R}\) is obviously 0, attained when x = 0. Zi Song Yeoh · 4 years, 3 months ago

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@Zi Song Yeoh i know, bt m talking about this case Abhimanyu Singla · 4 years, 3 months ago

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