Question

数学恒等变形请问怎样从4(a+b+c)^2-12(ab+bc+ac)变形到(a-b)^2+(b-c)^2+(a-c)^2

Answer 1

4(a+b+c)^2-12(ab+bc+ac)≥0

4（a^2+b^2+c^2+2ab+2ac+2bc）-12ab-12ac-12bc≥0

4a^2+4b^2+4c^2+8ab+8ac+8bc-12ab-12ac-12bc≥0

4a^2+4b^2+4c^2-4ab-4ac-4bc≥0

2a^2+2b^2+2c^2-2ab-2ac-2bc≥0

(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)≥0

(a-b)^2+(b-c)^2+(a-c)^2≥0