Áp dụng BĐT Bunhyaxcopki, ta có:
\(\left(x^2+y^2+z^2\right)\left(1^2+1^2+1^2\right)\ge\left(x+y+z\right)^2\)
\(\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge\left(\dfrac{3}{2}\right)^2\)
\(\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge\dfrac{9}{4}\)
\(\Leftrightarrow x^2+y^2+z^2\ge\dfrac{3}{4}\)
ủng hộ cách khác không xài bđt bunhia:
\(x^2+y^2+z^2\ge\dfrac{3}{4}\)
\(\Leftrightarrow x^2+y^2+z^2-x-y-z\ge\dfrac{3}{4}-\dfrac{3}{2}=-\dfrac{3}{4}\)
\(\Leftrightarrow x^2+y^2+z^2-x-y-z+\dfrac{3}{4}\ge0\)
\(\Leftrightarrow\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2-y+\dfrac{1}{4}\right)+\left(z^2-z+\dfrac{1}{4}\right)\ge0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2+\left(z-\dfrac{1}{2}\right)^2\ge0\)(luôn đúng \(\forall x+y+z=\dfrac{3}{2}\))
