Question

Chứng minh bất đẳng thức :

\(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\ge2xy\)

Answer 1

+) \(\left(x-y\right)^2\ge0\forall x;y\Leftrightarrow x^2-2xy+y^2\Leftrightarrow x^2+y^2\ge2xy\)

\(\Leftrightarrow2x^2+2y^2\ge x^2+y^2+2xy\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

\(\Rightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)(1)

+) \(x^2-2xy+y^2\ge0\forall x;y\Leftrightarrow x^2-2xy+y^2+4xy\ge4xy\)

\(\Leftrightarrow\left(x+y\right)^2\ge4xy\Rightarrow\frac{\left(x+y\right)^2}{2}\ge2xy\)(2)

Từ (1);(2) \(\Rightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\ge2xy\)(đpcm)