數學知識交流---求證

w⁴+ x⁴+ y⁴+ z⁴ - 4wxyz = 0( w⁴- 2w²x² + x⁴) + ( y⁴- 2y²z² + z⁴) + 2(w²x² + y²z² - 2wxyz) = 0(w² - x²)² + (y² - z²)² + 2(wx - yz)² = 0故 (w² - x²)² = 0 , (y² - z²)² = 0 及 2(wx - yz)² = 0即 w = x , y = z 及 wx = yz得 w = x = y = z故四邊相等 , 此四邊形為菱形或正方形。

By definition, w,x,y,z > 0, by AM-GM inequality,
=> [w^4+x^4+y^4+z^4] / 4 >= [(wxyz)^4]^(1/4)
=> w^4+x^4+y^4+z^4 >= 4wxyz
Given that w^4+x^4+y^4+z^4 = 4wxyz
Since the inequality hold when w^4 = x^4 = y^4 =z^4
Therefore, w=x=y=z
Hence the result follow.