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by quantskillsgmat » Wed Jan 04, 2012 12:13 am
a,b,c and d are 4 positive real numbers.and their product is 1.what is minimum value of (1+a)(1+b)(1+c)(1+d)
a)4
b)1
c)16
d)18
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by shankar.ashwin » Wed Jan 04, 2012 8:35 am
I really don't know if theres a proper method to do these problems you post. Most of them don't seem to be GMAT problems. Pls refrain posting random problems which are not within the scope of the exam.

I assume x=y=1/2 -> x+y =1

(5/2)^2 + (5/2)^2 = 12.5
quantskillsgmat wrote:If x and y both are positive and x+y=1.what is minimum value of (x+1/x)^2+ (y+1/y)^2
a)12
b)20
c)12.5
d)13.3
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by user123321 » Wed Jan 04, 2012 8:56 am
quantskillsgmat wrote:a,b,c and d are 4 positive real numbers.and their product is 1.what is minimum value of (1+a)(1+b)(1+c)(1+d)
a)4
b)1
c)16
d)18
since all are +ve. lets assume a=b=c=d=1, then you will get 16.
any other values apart from above like 1,1,2,1/2 will actually be above 16.
so min should be 16.

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by pemdas » Wed Jan 04, 2012 10:06 am
either 1/x or 1/y is close to 1
if 1/x is close to 1, then x is close to 1 and y is close to 0. One of fractions gets maximized, therefore take x=1/2 and y=1/2
(1/2+2)^2+(1/2+2)^2=50/4=12.5
c
quantskillsgmat wrote:If x and y both are positive and x+y=1.what is minimum value of (x+1/x)^2+ (y+1/y)^2
a)12
b)20
c)12.5
d)13.3
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by pemdas » Wed Jan 04, 2012 10:26 am
abcd must be four numbers, either whole or fractional. If whole their values is 1 and a=b=c=d
If fractional these numbers must cancel each other, e.g. 1/2*2*1/3*3.
In any case all fractional numbers added by one will be greater than a=b=c=d=1
hence, we minimize by 2^4=16
c

quantskillsgmat wrote:a,b,c and d are 4 positive real numbers.and their product is 1.what is minimum value of (1+a)(1+b)(1+c)(1+d)
a)4
b)1
c)16
d)18
Success doesn't come overnight!
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