Ta có: x3+y3+z3=3xyz
<=> x3+y3+z3-3xyz = 0
<=> (x+y+z)(x2+y2+z2-xy-yz-zx)=0
Do x,y,z dương nên x+y+z dương khi đó:
(x+y+z)(x2+y2+z2-xy-yz-zx)=0
=> x2+y2+z2-xy-yz-zx = 0
<=> 2x2-2y2-2z2-2xy-2yz-2xz = 0
=> (x-y)2+(y-z)2+(z-x)2 = 0
Do (x-y)2≥ 0 ; (y-z)2 ≥ 0 ; (z-x)2 ≥ 0 Nên: (x-y)2+(y-z)2+(z-x)2 = 0
\(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\)
Vậy nếu ba số x,y,z là ba số dương thỏa mãn
x3+y3+z3=3xyz thì x=y=z.
