回答：这是柯西－许瓦兹不等式(Cauchy-Schwarz Inequality).证：对于任意实变量t,考虑函数q(t) = E[(X+tY)^2]= E(Y^2)t^2 + 2E(XY)t + E(X^2).显然,对于一切实数t,q(t)≥0.这就意味着q(t)的判别式小于等于0,即4[E(XY...