\(x^2+y^2+z^2=xy+yz+zx\)
\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2zx\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Leftrightarrow x=y=z}\)
x^2+y^2+z^2=xy+yz+zx => x^2+y^2+z^2-xy-yz-zx = 0
<=> 2. (x^2+y^2+z^2-xy-yz-zx)=0
<=> (x^2-2xy+y^2) + (y^2-2yz+z^2)+(z^2-2zx+x^2)=0
<=> (x-y)^2 + (y-z)^2 + (z-x))^2 =0
Mà (x-y)^2, (y-z)^2, (z-x)^2 luôn >=0 với mọi x,y,z
=> x-y=y-z=z-x=0
=> x=y=z (ĐPCM)
