Consider X=1 and Y=1 now,

1/(sqrt(x+y)= 1/sqrt(2)= 1/1.41. <1 ----> 1

Now we need to find a value greater than 1

choice I = sqrt(x+y)/2x= sqrt(2)/2(1)=sqrt(2)/2=1/1.414 = same as eqn 1 so ruled out

choice II=[sqrt(x) +sqrt(y)]/(x+y)=[sqrt(1) + sqrt(1)]/2=2/2=1 which is > equation 1. So this option works.

choice III=[sqrt(x)-sqrt(y)]/(x+y)=0 which is < equation 1.

Hence answer is C.) II only

- Deepak

picking numbers doesn't work if you pick different numbers,
eg if you choose x=1 and y=2.
In this case option I will be larger.
The reason why II is the answer is because it the only one that must be larger and that is because:

1/sqrt(x+y)= sqrt(x+y)/(x+y)
the denominator of this fraction is the same as the one in II so the fraction with the smaller numerator will be smaller and sqrt(x+y) is always smaller than sqrt(x)+sqrt(y)