f(x)>0证∫abf(x)*∫ab1/f(x)≥(b-a)^2

Question

f(x)>0 证∫abf(x)*∫ab1/f(x)≥(b-a)^2

Answer 1

令f(x)=(∫b a f(t)dt ) x^2 -(2∫b a 1dt)x +(∫b a 1/f(t)dt),则：f(x)=∫b a f(t) x^2 dt -2∫b a xdt +∫b a 1/f(t)dt =∫b a [f(t) x^2 -2x +1/f(t)]dt=∫b a {[f(t)^0.5 x -1/f(t)^0.5]^2}dt ≥0故这个关...