Cho (x – y)² + (y – z)² + (z – x)² = 4(x² + y² + z² – xy – yz – zx). Chứng minh x = y = z.

Ta có (x – y)² + (y – z)² + (z – x)² = 4(x² + y² + z² – xy – yz – zx).

⇔ x² – 2xy + y² + y² – 2yz + z² + z² – 2zx + x² = 4(x² + y² + z² – xy – yz – zx).

⇔ 2x² + 2y² + 2z² – 2xy – 2yz – 2zx = 4(x² + y² + z² – xy – yz – zx).

⇔ 2(x² + y² + z² – xy – yz – zx) = 4(x² + y² + z² – xy – yz – zx).

⇔ 2(x² + y² + z² – xy – yz – zx) = 0.

⇔ x² – 2xy + y² + y² – 2yz + z² + z² – 2zx + x² = 0.

⇔ (x – y)² + (y – z)² + (z – x)² = 0.

$\iff \left\{\begin{array}{l}x-y=0\\ y-z=0\\ z-x=0\end{array}\right.\iff \left\{\begin{array}{l}x=y\\ y=z\\ z=x\end{array}\right.\iff x=y=z$

Vậy ta có điều phải chứng minh.