不等式的证明......

题目是a和b都是正数吧。

1/a+1/b
=1/4 * (a+b)(1/a+1/b)
=1/4 * (2+a/b+b/a)
>=1/4 * (2 + 2根号(a/b*b/a))
=1/4 * (2+2)
=1

(a+b)(1/a+1/b)-4
=(a+b)[(a+b)/ab]-4
=(a+b)^2/ab-4
=(a^2+2ab+b^2-4ab)/ab
=(a-b)^2/ab
∵a>0,b>0
∴(a-b)^2/ab≥0
∴(a+b)(1/a+1/b)-4≥0
∵a+b=4
∴4(1/a+1/b)≥4
∴1/a+1/b≥1

a+b=4大于等于2倍根号ab，所以ab小于等于4
(1/a)+(1/b)=(a+b)/ab=4/ab≥1