Question

Which one of the following hold good?

A
abcd>81(sa)(sb)(sc)(sd)
where 3s=a+b+c+d
B
(sa)(sb)(sc)(sd)>81abcd
where s=a+b+c+d
C
If a+b+c=1, then
827abc>{1a1}{1b1}{1c1}>8
D
If x+y+z=a then
827a3(ax)(ay)(az)8xyz

Solution

The correct options are A abcd>81(s−a)(s−b)(s−c)(s−d) where 3s=a+b+c+d B (s−a)(s−b)(s−c)(s−d)>81abcd where s=a+b+c+d C If a+b+c=1, then 827abc>{1a−1}{1b−1}{1c−1}>8 D If x+y+z=a then 827a3≥(a−x)(a−y)(a−z)≥8xyz(a), (b), (c), (d) (a) a=3x−b−c−d or =(s−b)+(s−c)+(s−d) >3[(s−b)(s−c)(s−d)]13, ∵A.M.>G.M. Similarly write other inequalities for b,c and d and then multiply (b) s−a=b+c+d>3(bcd)13 Similarly write other inequalities for (s-b)(s-c) and (s-d) and multiply. (c) On multiplying by abc, we have to prove that 827>(1−a)(1−b)(1−c)>8abc Now, (1−a)+(1−b)+(1−c)3>[(1−a)(1−b)(1−c)]13 or 3−a+b+c3>[(1−a)(1−b)(1−c)]13 or (23)3>(1−a)(1−b)(1−c) ∵a+b+c=1 which proves the 1st part. The 2nd part of the inequality has been proved.

Suggest corrections