Câu hỏi của Phan Văn Tấn

Bài 7:Ta có: \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\forall x,y,z\) =>\(2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\ge0\forall x,y,z\) =>\(\left(x^2+y^2+z^2\right)\ge xy+yz+xz\) =>\(x^2+y^2+z^2+2\left(xy+yz+xz\right)\ge3\left(xy+yz+xz\right)\) =>\(\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\) với mọi x,y,z=>\(3\left(xy+yz+xz\right)\le3^2=9\) =>xy+yz+xz \(2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\ge0\forall x,y,z\) =>\(\left(x^2+y^2+z^2\right)\ge xy+yz+xz\) =>\(x^2+y^2+z^2+2\left(xy+yz+xz\right)\ge3\left(xy+yz+xz\right)\) =>\(\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\) với mọi x,y,z=>\(3\left(xy+yz+xz\right)\le3^2=9\) =>xy+yz+xz<=3 Dấu '=' xảy ra khi x=y=zmà x+y+z=3nên \(x=y=z=\frac33=1\)