Question

若a＞0,b＞0,1／a+3／b=1,则a+2b的最小值是 求详解！！！！！！！！！！！！

Answer 1

(a+2b)*1=(a+2b)(1/a+3/b)
=1+2b/a+3a/b+6
>=7+2√6
2b/a=3a/b取等
你如果会柯西不等式，也可以直接用柯西
(a+2b)(1/a+3/b)>=(1+√6)^2=7+2√6

Answer 2

1/a+3/b=1 => b=3a/(a-1), a>1, b>3
a+2b=a+6a/(a-1) = a+6+6/(a-1) =7 +(a-1)+6/(a-1)
(a-1)+6/(a-1) ≥ 2√6
a+2b ≥ 7+2√6
a+2b的最小值是7+2√6

Answer 3

解：
令y = a + 2b
a = b/(b-3)
则：y = b/(b-3) + 2b
所以：2b^2 - (y + 5)b + 3y = 0
因此：(y + 5)^2 - 24y ≥ 0
y ≥ 7 + 2√6 或者y≤7-2√6
a + 2b ≥ 7 + 2√6