\(m^2+n^2+2\ge2\left(m+n\right)\\ \Leftrightarrow\left(m^2+2m+1\right)+\left(n^2+2n+1\right)\ge0\\ \Leftrightarrow\left(m+1\right)^2+\left(n+1\right)^2\ge0\forall m,n\)
m2 + n2 + 2 ≥ 2 (m + n )
⇔m2+n2+2-2m-2n≥0
⇔m2+n2+1+1-2m-2n≥0
⇔m2-2m+1+n2+2n+1≥0
⇔(m-1)2+(n-1)2≥0 (luôn đúng)
`m^2+n^2+2>=2(m+n)`
`<=>m^2+n^2+2>=2m+2n`
`<=>m^2+n^2+2-2m-2n>=0`
`<=>(m^2-2m+1)+(n^2-2n+1)>=0`
`<=>(m-1)^2+(n-1)^2>=0`
Ta thấy : `(m-1)^2>=0,∀m`
`(n-1)^2>=0,∀n`
`=>(m-1)^2+(n-1)^2>=0 \ ( ∀m,n)`