由a,b大于0且2a+b=1,
有a=(1-b)/2,b=1-2a,0t=1/a+1/2b=2/(1-b)+1/2b==(1+b)/2b*(1-b)
1/t=2b*(1-b)/(1+b)
=2(b-b²)/(1+b)
=- 2(b²+2b+1-3b-3+2)/(1+b)
=- 2【(b+1)²-3(b+1)+2】/(1+b)
= -2【(b+1)+2/(b+1)-3】
当(b+1)=2/(b+1) 时 1/t值最大 为6-4√2
t=1/(6-4√2) =(6+4√2)/4=(3+2√2)/2 =3/2 +2√2
1/a+1/2b
=(1/a+1/2b)*1
=(1/a+1/2b)(2a+b)
=2+1/2+a/b+b/a
>=5/2+2
=7/2