We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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