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

Mathematics
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i have no idea how to do this
looks like you need to rearrange. you can get three squares on the lhs.
yes...> or = to 0.

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(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)
OH MY WORD! THANK YOU!
your welcome!
that answer is incorrect.

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