\[(a+b+c)^{2}=1\]
\[a^{2}+b^{2}+c^{2}+2ab+2bc+2ca=1\]
\[a^{2}+b^{2}+c^{2}-ab-bc-ca+3ab+3bc+3ca=1\]
\[\frac {1}{2}[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]+3(ab+bc+ca)=1\]
\[3(ab+bc+ca)\] 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
·
3 years, 12 months ago

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
·
3 years, 12 months ago

## Comments

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TopNewest\[(a+b+c)^{2}=1\] \[a^{2}+b^{2}+c^{2}+2ab+2bc+2ca=1\] \[a^{2}+b^{2}+c^{2}-ab-bc-ca+3ab+3bc+3ca=1\] \[\frac {1}{2}[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]+3(ab+bc+ca)=1\] \[3(ab+bc+ca)\] 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 · 3 years, 12 months ago

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– Rohan Rao · 3 years, 12 months ago

Nice proof!Log in to reply

AM-GM kills it – Taufiq Hakim Rac'madhan · 3 years, 11 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 · 3 years, 12 months ago

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– Adhiraj Mandal · 3 years, 12 months ago

The result also follows from the Rearrangement-InequalityLog in to reply

solution? – Shubham Bagrecha · 3 years, 12 months ago

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You can also use LaGrange Multipliers to solve for the maximum value right? – Eric Wu · 3 years, 11 months ago

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1/3 – Abhimanyu Singla · 3 years, 12 months ago

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1/3 is the maximum – Ed Mañalac · 3 years, 12 months ago

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0.33 – Nishit Goyal · 3 years, 12 months ago

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– Zi Song Yeoh · 3 years, 12 months ago

0.333333... To be precise (or simply 1/3, you meant 33/100)Log in to reply

The max. value of ab + bc + ca is 0. – Sarthak Nandan · 3 years, 12 months ago

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– Abhimanyu Singla · 3 years, 11 months ago

that's wrongLog in to reply

– Abhimanyu Singla · 3 years, 12 months ago

maximum is never 0Log in to reply

– Zi Song Yeoh · 3 years, 12 months ago

Not always trueLog in to reply

– Zi Song Yeoh · 3 years, 12 months ago

E.g. Maximum of -x^2 where \(x \in \mathbb{R}\) is obviously 0, attained when x = 0.Log in to reply

– Abhimanyu Singla · 3 years, 11 months ago

i know, bt m talking about this caseLog in to reply