1.
Since a, b, c, d>1, ab+1>a+b, cd+1>c+d
2ab+2>ab+ab+1, 2cd+2>cd+c+d+1
(ab+a+b+1)(cd+c+d+1)<(2ab+2)(2cd+2)<8abcd+8
so, (a+1)(b+1)(c+1)(d+1)<8(abcd+1)
2.
Assume that a>=b>=c, then 1/(b+c)>=1/(a+c)>=1/(a+b)
Accroding to the rearrangement inequality,
a/(b+c)+b/(a+c)+c/(a+b)>=a/(a+c)+b/(a+b)+c/(b+c)
and a/(b+c)+b/(a+c)+c/(a+b)>=a/(a+b)+b/(b+c)+c/(a+c)
Add the two inequalities above, we get
2[a/(b+c)+b/(a+c)+c/(a+b)]>=1+1+1=3
=> c/(a+b)+b/(a+c)+a/(b+c)≧3/2