Question

求极值问题

f(x)=3x+a,g(x)=3x+2a在(b,c)上都有零点，求(a2+2ab+2ac+4bc)/(b2-2bc+c2)最小值

Answer 1

a2+2ab+2ac+4bc = (a+2b))(a+2c)

如图，红线为a>0时，绿线为a<0时

a>0时，b < -2a/3, c>-a/3

(b-c)^2 >a^2/9

a+2b < -a/3

a+2c > a/3

(a+2b)(a+2c) < -a^2/9

(a2+2ab+2ac+4bc)/(b2-2bc+c2) 的最小值 = -1