Vicky wrote:

Answer: C (16)

We use the property that:

Arithemetic Mean of two numbers is g.t.e.q Geometric mean.

g.t.e.q = greater than or equal to.

Thus (1+a)/2 >= sqrt(a)

(1+b)/2 >= sqrt(b)

--------

(1+d)/2 >= sqrt(d)

Multiplying above equations (we can mutilply them without chanding sign of inequality because it is given that a,b,c & d are postive real num's)

we get (1 + a) (1 + b) (1 + c) (1 + d)/2^4 >= sqrt (abcd)

Thus minimum is 16.

-Vicks

ps: let me know if u know a faster way to solve this.

Vicky

Unless Akamai or Stolyar have a good idea,yours is really the best analytical approach...very convincing and leaves no room for error.

as for me, I did it this way.

Since the answer choices are integers..i find it very likely that a,b,c,d will be integers

As a check...For example let a=0.5 ,b=2 ,c=2 , d=0.5

a*b*c*d=0.5*2*2*0.5 =1 ...here we have integers and non integers..and we satsify the abcd =1

But (1+a)(1+b)(1+c)(1+d)= 1.5*3*3*1.5 =>> not an integer ...

Another one

3*1/3*2*1/2 =1

but 4*4/3*3*3/2 ==>>not an integer

So lets work with integers first

The answer cannot be 1 as the product is 1 + something ...and the something is positive

4= 1*1*2*2 ..compare with (1+a) (1+b)(1+c)(1+d)

We get a=0 and b =0 ..not possible..as abcd = 1....doesnt satisfy

18= 1*2*3*3 again comparison gives a=0...not possible

16= 2*2*2*2 ...comparison gives a=1,b=1,c=1,d=1...its the only one that holds true.

thanks

praetorian