已知a,b為正實數,求證:(a+b)×(1/a+1/b)≥4

已知a,b為正實數,求證:(a+b)×(1/a+1/b)≥4

證:
(a+b)(1/a+1/b)
=1+a/b+b/a+1
=2+a/b+b/a
a>0,b>0
由均值不等式,得:a/b+b/a≥2(當a/b=b/a時,即a=b時,取等號)
2+a/b+b/a≥2+2=4
(a+b)(1/a+1/b)≥4