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.

FencingKat Nov 8, 2018

#1**+2 **

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

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

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

.Rom Nov 8, 2018