Question

cho a,b,c tuỳ ý . Chứng minh a²+b²+c²≥ ab+bc+ac

Giúp mik vs ạ

Answer 1

Có: `a^2+b^2+c^2 >= ab+bc+ac`

`<=>2a^2+2b^2+2c^2 >= 2ab+2bc+2ac`

`<=>a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2 >= 0`

`<=>(a-b)^2+(b-c)^2+(c-a)^2 >= 0` (LĐ `AA a,b,c`)

Answer 2

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

\(2a^2+2b^2+2c^2\ge2ab+2bc+2ac\)

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

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\left(lđ\right)\)