a+2b=2,a>0,b>0,则1/a+1/b最小值是多少
问题描述：
a+2b=2,a>0,b>0,则1/a+1/b最小值是多少
答
1/a+1/b
=1/2×(2/a+2/b)
=1/2×[(a+2b)/a+(a+2b)/b]
=1/2×(3+2b/a+a/b)
≥1/2×[3+2√(2b/a×a/b)]
=1/2×[3+2√2]
=(3+2√2)/2
1/a+1/b最小值是(3+2√2)/2