设f(x)=√x p,q>0,且p+q=1,求证pf(x)+qf(x)

Question

如题所述

Answer 1

f(x)=√x
pf(x1)+qf(x2)=p√x1+q√x2
f(px1+qx2)=√(px1+qx2)
p,q大于0,
则，[pf(x1)+qf(x2)]^2-[f(px1+qx2)]^2=
p^2x1+q^2x2+2pq√x1x2-(px1+qx2)
=(p^2-p)x1+(q^2-q)x2+2pq√x1x2
=p(p-1)x1+q(q-1)x2+2pq√x1x2
p+q=1,则
p(p-1)x1+q(q-1)x2+2pq√x1x2
=-(pqx1+pqx2)+2pq√x1x2
由重要不等式得：
(pqx1+pqx2)≥2√pqx1pqx2=2pq√x1x2
所以-(pqx1+pqx2)+2pq√x1x2≤0
所以[pf(x1)+qf(x2)]^2-[f(px1+qx2)]^2≤0
所以[pf(x1)+qf(x2)]≤[f(px1+qx2)]^2

Answer 2

题目打错了,应该为pf(x1)+qf(x2)0
显然成立