\(\dfrac{x+y}{2}\le\dfrac{\left|x\right|+\left|y\right|}{2}\le\sqrt{\dfrac{\left(\left|x\right|+\left|y\right|\right)^2}{4}}\le\sqrt{\dfrac{x^2+y^2-2\left|x\right|\left|y\right|}{4}\le\sqrt{\dfrac{x^2+y^2}{4}}}\le\sqrt{\dfrac{x^2+y^2}{2}}\)
*Với a + b < 0 thì bài toán luôn đúng
*Với a + b > 0 . Bình phương 2 vế ta đc
\(\dfrac{\left(a+b\right)^2}{4}\le\dfrac{a^2+b^2}{2}\)
\(\Leftrightarrow\dfrac{a^2+2ab+b^2}{2}\le a^2+b^2\)
\(\Leftrightarrow a^2+2ab+b^2\le a^2+b^2\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\) ( Luôn đúng)
Dấu "=" khi a = b