Question

Chứng minh rằng nếu x-y+z=0 thì xy+yz+xz\(\ge\)

Giúp với!!!

Answer 1

\(\ge\)0 nhá

Answer 2

Ta có: \(x-y+z=0\)
    \(\Rightarrow\left(x-y+z\right)^2=0 \)
  \(\Rightarrow\left(x-y+z\right).\left(x-y+z\right)=0\)
   \(\Rightarrow x\left(x-y+z\right)-y\left(x-y+z\right)+z\left(x-y+z\right)=0\)
   \(\Rightarrow x^2-xy+xz-xy+y^2-yz+xz-yz+z^2=0\)
  \(\Rightarrow x^2+y^2+z^2=xy+xy+yz+yz-xz-xz\)
   \(\Rightarrow x^2+y^2+z^2=2xy+2yz-2xz\)
   \(\Rightarrow x^2+y^2-z^2=2\left(xy+yz-xz\right)\)
Mà: \(x^2+y^2-z^2\ge0\)
\(\Rightarrow2\left(xy+yz-xz\right)\ge0\)
\(\Rightarrow xy+yz-xz\ge0\)(đpcm)
   Vậy: \(xy+yz-xz\ge0\)