Question

已知正实数abcd满足a+b=1,c+d=1 则1/（abc）+1/d最小值

Answer 1

ab<=(a+b)^2/4=1/4,a=b=1/2时取等号，d=1-c
∴w=1/（abc）+1/d>=4/c+1/(1-c)
=[4(1-c)+c]/[c(1-c)]
=(4-3c)/(c-c^2),记为f(c),0<c<1,
f'(c)=[-3(c-c^2)-(1-2c)(4-3c)]/(c-c^2)^2
=-(3c^2-11c+4)/(c-c^2)^2
=-3[c-(11-√73)/6][c-(11+√73)/6]/(c-c^2)^2,
0<c<c-(11-√73)/6时f'(c)<0,f(c)是减函数；c-(11-√73)/6<c<1时f'(c)>0,f(c)是增函数：
∴f(c)的最小值=f[(11-√73)/6]=(49+5√73)/8,为所求。

Answer 2

ab≤1/4
所以1/(abc)≥4/c
所以1/(abc)+1/(1-c)≥4/c+1/(1-c)＝(4-3c)/[(1-c)c]
设f(c)＝(4-3c)/[c(1-c)]
设4-3c＝t属于(1,4)
所以c＝(4-t)/3
f(t)＝9t/(4-t)(t-1)
＝-9/(t-5+4/t)≥-9/(4-5)＝9
当t＝2，即c＝2/3，d＝1/3取得最小值
所以最小值9

追答

a=b＝1/2,c＝2/3,d＝1/3

Answer 3

设a≥b≥c 则a2+b2+c2≥ac+b2+ca 且a2+b2+c2≥ab+bc+ca 2(a2+b2+c2) ≥ac+b2+ca+ab+bc+ca =3ac+b(a+b+c) ≥3ac+ab2c =ac(3+b2) ≥ac(2√3b) =2√3abc 所以a2+b2+c2≥√3abc