Question

m² + n² + 2 ≥ 2 (m + n )

Answer 1

\(m^2+n^2+2\ge2\left(m+n\right)\\ \Leftrightarrow\left(m^2+2m+1\right)+\left(n^2+2n+1\right)\ge0\\ \Leftrightarrow\left(m+1\right)^2+\left(n+1\right)^2\ge0\forall m,n\)

Answer 2

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)

Answer 3

`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)`