4(a+b+c)/3 = 1/3 >= (abc)1/3
=> abc <= 1/27
Also using AM >= HM v get 1/a + 1/b + 1/c >= 9
Given exprsn = 1 + 1/a+1/b+1/c +2/abc >= 1 + 9 + 54 >= 64
Hence proved
8 Answers
4
62One simple way will be fix a and then prove that b=c (ie for any given a, the expression will be minimum if b=c
then prove a=b if c is fixed and a=c if b is fixed..
hence for maximum, a=b=c
[1]
1my post got deleted
i think a,b,c should be positive
inequality is equivalent to
(a+1)(b+1)(c+1)\ge 64abc
set 1=a+b+c
(a+a+b+c)(b+a+b+c)(c+a+b+c)\ge 64abc
which is true by AM-GM on individual brackets
1i think my method is correct......
SOLUTION:
AM>=HM
or (a+b+c)/3 >= 3/(1/a +1/b + 1/c)
or (1/a +1/b + 1/c) >= 9/(a+b+c) = 9
now takinng limiting case,
max. value of 1/a = 1/b = 1/c= 3.
so, 1+ 1/a =1+ 1/b =1+ 1/c=4.
(1+ 1/a) (1+ 1/b)(1+ 1/c)>= 4^3= 64
HENCE PROVED...
let nishant bhaiyya check this...
62@ arpan...
why do you have a maxima at the limiting case and not a minima?
3411+\frac{1}{a} = 1 + \frac{1}{3a} + \frac{1}{3a} +\frac{1}{3a} \ge 4 \left(\frac{1}{27abc}\right)^{\frac{1}{4}}
Multiplying three similar inequalities, we get
\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) = \ge 64 \left(\frac{1}{27abc}\right)^{\frac{3}{4}}
Using AM-GM inequality, we have
a+b+c =1 \Rightarrow \frac{1}{3} \ge (abc)^{\frac{1}{3}} \Rightarrow \frac{1}{27abc} \ge 1
Thus LHS ≥64
3412nd one is fairly obvious with factorisation and AM-GM on one of the factors
