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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
3 years, 10 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. $ab+bc+ca=\frac {1}{3}$ · 3 years, 10 months ago

Nice proof! · 3 years, 10 months ago

AM-GM kills it · 3 years, 9 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 · 3 years, 10 months ago

The result also follows from the Rearrangement-Inequality · 3 years, 10 months ago

solution? · 3 years, 10 months ago

You can also use LaGrange Multipliers to solve for the maximum value right? · 3 years, 9 months ago

1/3 · 3 years, 10 months ago

1/3 is the maximum · 3 years, 10 months ago

0.33 · 3 years, 10 months ago

0.333333... To be precise (or simply 1/3, you meant 33/100) · 3 years, 10 months ago

The max. value of ab + bc + ca is 0. · 3 years, 10 months ago

that's wrong · 3 years, 9 months ago

maximum is never 0 · 3 years, 10 months ago

Not always true · 3 years, 10 months ago

E.g. Maximum of -x^2 where $$x \in \mathbb{R}$$ is obviously 0, attained when x = 0. · 3 years, 10 months ago