設三角形的邊長為a,b,c 證明 a^2b(a-b)+b^2c(b-c)+c^2a(c-a) >= 0
假設考慮一個函數 f(x) = x^2b(x-b)+b^2c(b-c)+c^2x(c-x) 其中 x>=b, b>=c
因為 f(b) >=0 且 f'(x) >=0 在 x>=b
故得證
請問這樣的證明方式有錯嗎?
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