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

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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## 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. \[ab+bc+ca=\frac {1}{3}\]

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Nice proof!

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solution?

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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

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The result also follows from the Rearrangement-Inequality

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AM-GM kills it

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1/3 is the maximum

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You can also use LaGrange Multipliers to solve for the maximum value right?

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0.33

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

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1/3

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The max. value of ab + bc + ca is 0.

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that's wrong

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maximum is never 0

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Not always true

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