a) Ta có:
\(x^2+y^2\ge2xy\Leftrightarrow x^2+y^2-2xy\ge0\Leftrightarrow\left(x-y\right)^2\ge0\), đúng \(\forall x,y\in R\)
Dấu "=" xảy ra \(\Leftrightarrow x-y=0\Leftrightarrow x=y\)
Vậy ta có đpcm
a) \(\Leftrightarrow x^2+y^2-2xy\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\left(đúng\right)\)
b) \(\Leftrightarrow2x^2+2y^2\ge x^2+2xy+y^2\)
\(\Leftrightarrow x^2+y^2\ge2xy\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\left(đúng\right)\)
c) \(\Leftrightarrow\dfrac{2x^2+2y^2}{4}\ge\dfrac{x^2+2xy+y^2}{4}\)
\(\Leftrightarrow2x^2+2y^2\ge x^2+2xy+y^2\)
\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)( đúng do đã CM ở câu b)
b)Từ câu a ta có:
\(x^2+y^2\ge2xy\Leftrightarrow x^2+y^2+x^2+y^2\ge x^2+y^2+2xy\)
\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\left(đpcm\right)\)
Dấu "=" xảy ra\(\Leftrightarrow x=y\)
c)Ta có:
\(\dfrac{x^2+y^2}{2}=\dfrac{2\left(x^2+y^2\right)}{4}\ge\dfrac{\left(x+y\right)^2}{4}=\left(\dfrac{x+y}{2}\right)^2\)
Vậy ta có đpcm.
Dấu "=" xảy ra \(\Leftrightarrow x=y\)
.
BC.
