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Question

Which one of the following hold good?


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

Solution

The correct options are
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
(a), (b), (c), (d)
(a) a=3xbcd
or =(sb)+(sc)+(sd)
>3[(sb)(sc)(sd)]13, A.M.>G.M.

Similarly write other inequalities for b,c and d and then multiply

(b) sa=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>(1a)(1b)(1c)>8abc
Now, (1a)+(1b)+(1c)3>[(1a)(1b)(1c)]13
or 3a+b+c3>[(1a)(1b)(1c)]13
or (23)3>(1a)(1b)(1c)
a+b+c=1
which proves the 1st part.
The 2nd part of the inequality has been proved.

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