已知正项等比数列an满足:a3=a2+2a1,若存在两项am,an使得根号下aman=4a1,1/m+4/n的最小值为

a(n) = aq^(n-1),a>0,q>0.
a(3)=aq^2 = a(2)+2a(1) = aq+2a,
q^2 = q + 2,
0 = q^2 - q - 2 = (q-2)(q+1),q=2.
a(n) = a*2^(n-1).
a(m)=a*2^(m-1).
4a(1) = 4a = [a(m)a(n)]^(1/2) = a*2^[(m+n-2)/2],
4 = 2^[(m+n-2)/2] = 2^2,
2 = (m+n-2)/2,
4 = m+n-2,
m=6-n.
1m+n=6, 1=(m+n)/6.

1/m + 4/n = [1/m + 4/n][(m+n)/6] = [m/m + 4m/n + n/m + 4n/n]/6 = (1/6)(5+ n/m + 4m/n)

m>=1, n>=1.
最后一步利用不等式 a>0,b>0时，a+b >= 2(ab)^(1/2),
{因为 0<= [a^(1/2) - b^(1/2)]^2 = a +b - 2(ab)^(1/2)}

所以，n/m + 4m/n >= 2[(n/m)(4m/n)]^(1/2) = 2*2 = 4.

1/m + 4/n = (1/6)[5 + n/m + 4m/n] >= (1/6)[5 + 4] = 9/6 = 3/2.