Pure maths inequality

The question is strange. If we only use the AM>=GM inequality, the result would be
1/[a^3(b+c)]+1/[b^3(a+c)]+1/[c^3(a+b)]
>=3{1/[a^3(b+c)][b^3(a+c)][c^3(a+b)]}^(1/3)
=3{1/[(b+c)(a+c)(a+b)]}^(1/3)
So consider [(b+c)(a+c)(a+b)]}^(1/3)
<=(b+c+c+a+a+b)/3
=2(a+b+c)/3
So 1/[a^3(b+c)]+1/[b^3(a+c)]+1/[c^3(a+b)]
>=3*3/2(a+b+c)
=9/[2(a+b+c)]
And that's all since (a+b+c)/3 is larger than 1 not smaller than 1
The correct approach is
Let S=1/[a^3(b+c)]+1/[b^3(a+c)]+1/[c^3(a+b)]
Define x=1/a,y=1/b,z=1/c and T=x+y+z
Then 1/[a^3(b+c)]=x^3/[(1/y)+(1/z)]=x^2/(T-x)=(T^2-T^2-x^2)/(T-x)
=T^2/(T-x)-T-x
Similar hor the other two
So S=T^2[1/(T-x)+1/(T-y)+1/(T-z)]-4T
Using AM-GM-HM inequality
S
=T^2[1/(T-x)+1/(T-y)+1/(T-z)]-4T
>=T^2[9/(T-x)+(T-y)+(T-z)]-4T
=T/2
=(x+y+z)/2
>=3/2 (since xyz=1)
Other method
Since 1/[a^3(b+c)]=x^2/(y+z)
S=x^2/(y+z)+y^2/(x+z)+z^2/(x+y)
By Cauchy-Schwarz inequality
[(y+z)+(x+z)+(x+y)]S>=(x+y+z)^2
That is S>=(x+y+z)/2

Try to directly applying AM >= GM:
1/[a^3(b+c)]+1/[b^3(a+c)]+1/[c^3(a+b)]>=3/[a^3(b+c)b^3(a+c)c^3(a+b)]^(1/3)
= 3/[(b+c)(a+c)(a+b)]^(1/3)
Then try to prove 1/[(b+c)(a+c)(a+b)]^(1/3) >= 1/2 or alternatively, (b+c)(a+c)(a+b)<= 8

yea, so can you help me to solve it?

2009-07-16 16:19:34 補充：
i have done that before.
however, (a+b)(b+c)(a+c) is always >=8abc
it can be done by directly using am>gm

in this question, (a+b)(b+c)(a+c)>=8abc=8

Applying A.M. >= G.M. again:
1/a5/2 + 1/b5/2 + 1/c5/2 <= 3(a5/2b5/2c5/2)1/3
The inequality sign is wrong.