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anonymous
 5 years ago
let a, b, c be real numbers prove:
a^2+B^2+c^2+3>(or = to) 2(a+b+c)
anonymous
 5 years ago
let a, b, c be real numbers prove: a^2+B^2+c^2+3>(or = to) 2(a+b+c)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have no idea how to do this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0looks like you need to rearrange. you can get three squares on the lhs.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(a + b + c)^2 = a^2 + ab + ac + ab + b^2 + bc + ac + bc + c^2 = a^2 + b^2 + c^2 + 2ab + 2 ac + 2bc ===> a^2 + b^2 + c^2 = (a + b + c)^2  (2ab + 2 ac + 2bc) ===> a^2 + b^2 + c^2 = (a + b + c)^2  2(ab + ac + bc) ===> (a^2 + b^2 + c^2) = (a + b + c)^2  2(ab + ac + bc) For all real a, b and c, a^2 + b^2 + c^2 ≥ 0 and (a + b + c)^2 ≥ 0. For all real a, b and c, a^2 + b^2 + c^2 ≥ 0 and (a + b + c) ≥ 0. For all real a, b and c, a^2 + b^2 + c^2 ≥ 0 and 2(a + b + c) ≥ 0. Therefore, a^2 + b^2 + c^2 ≥ 2(a + b + c)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OH MY WORD! THANK YOU!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that answer is incorrect.
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