Ta cm 1 bđt phụ sau:
\(xy+yz+xz\le\frac{\left(x+y+z\right)^2}{3}\)
Thật vậy: \(bdt\Leftrightarrow3\left(xy+yz+xz\right)\le\left(x+y+z\right)^2\)
\(\Leftrightarrow xy+yz+xz\le x^2+y^2+z^2\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\) (đúng)
\(\Rightarrow xy+yz+xz\le\frac{\left(x+y+z\right)^2}{3}=3\)
\("="\Leftrightarrow x=y=z=1\)