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Let \(a, b, c, d, e, f\) be nonnegative real numbers such that \(a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6\) and \(ab + cd + ef = 3\). What is the maximum value of \(a+b+c+d+e+f\)?

 

Hi! Thanks for the help!

 

Please provide a full solution so I may understand how to do this problem step by step.

 

I think that you have to use Cauchy-Schawz in order to find this but I'm not sure. 

 Nov 8, 2018
 #1
avatar+3589 
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\(\text{let }u = (a,c,e),~v = (b,d,f)\\ \text{we are given that }\\ u\cdot u + v\cdot v = 6\\ u\cdot v = 3\\ \text{and we are asked for max of }(u+v)\cdot (1,1,1)\)

 

\(\text{By Cauchy-Schwarz}\\ |(u+v)\cdot(1,1,1)|^2 \leq (u+v)\cdot (u+v) \times (1,1,1)\cdot (1,1,1) = \\ (u\cdot u + v\cdot v + 2u\cdot v)(3) = \\ (6+2(3))(3) = 36 \)

 

\(\text{so }(u+v)\cdot (1,1,1)\leq 6 \\ \text{with the maximum occurring at 6}\\ \text{thus the maximum of }a+b+c+d+e+f = 6\)

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 Nov 8, 2018
 #2
avatar+39 
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Thank you!

 Nov 8, 2018

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