Question

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

Answer 1

i have no idea how to do this

Answer 2

looks like you need to rearrange. you can get three squares on the lhs.

Answer 3

yes...> or = to 0.

Answer 4

(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)

Answer 5

OH MY WORD! THANK YOU!

Answer 6

your welcome!

Answer 7

that answer is incorrect.