Question

已知a.b.c.满足a^2+b^2=1;b^2+c^2=2;c^2+a^2=2 求ab+bc+ca的最小值

Answer 1

ab+bc+ca = 1/2{(a+b+c)^2-(a^2+b^2+c^2)} = 1/4{2(a+b+c)^2-(a^2+b^2)-(b^2+c^2)-(c^2+a^2)}

= 1/2(a+b+c)^2 - 1/4{(a^2+b^2)+(b^2+c^2)+c^2-a^2)}

= 1/2(a+b+c)^2 - 1/4(1+2+2)

= 1/2(a+b+c)^2 - 5/4

1/2(a+b+c)^2≥0

1/2(a+b+c)^2 - 5/4 ≥-5/4

ab+bc+ca的最小值-5/4