Question

Chứng minh bất đẳng thức: \(a^2+b^2+2\ge2\left(a+b\right)\)

Answer 1

ta có : \(\left(a-1\right)^2\ge0\forall a\Rightarrow a^2-2a+1\ge0\Rightarrow a^2+1\ge2a\left(1\right)\)

\(\left(b-1\right)^2\ge0\forall b\Rightarrow b^2+1\ge2b\left(2\right)\)

Lấy (1)+(2) ta có :  \(a^2+1+b^2+1\ge2a+2b\forall a,b\)

\(\Rightarrow a^2+b^2+2\ge2\left(a+b\right)\forall a,b\)

Answer 2

Theo BĐT AM - GM :

\(a^2+1\ge2\sqrt{a^2}=2\left|a\right|=2a\)

\(b^2+1\ge2\sqrt{b^2}=2\left|b\right|\ge2b\)

Khi đó ta có đpcm