Question

Chứng minh BĐT sau:

\(a^2+b^2+c^2\ge a.\left(b+c\right)\)

Ai giúp đc cám ơn nha!<3

Answer 1

\(a^2+b^2+c^2\ge a\left(b+c\right)\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+ac\)

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2ac\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac\ge0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+b^2+c^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+b^2+c^2\ge0\)(luôn đúng \(\forall a;b\))

Vậy \(a^2+b^2+c^2\ge a\left(b+c\right)\)

Answer 2

Thanks you!<3