Question

Assuring that x, yare positive
real numbers, simplify:
: (x*-2/3 . y*-1/2)*2 and √X*-2 Y*3​

Answer 1

Answer:

voodoochild wrote:

Experts,

I liked Walker's method. I chose X = y =1 and got 'none'. However, can we solve this question algebraically?

[Note: x and y are +ve; hence both the sides of inequalities are positive. There shouldn't be any harm in squaring]

#1 - This is straightforward.

For #2 - here's what I did: (Sqrt (x) + Sqrt (y))/2 > 1/(Sqrt(x+y))

Square both sides:

x + y + 2*sqrt(xy) > 24/(x+y)

OR

(x+y)^2 + 2(x+y)Sqrt (xy) > 4

This seems to hold good only when (x+y)^2 is greater than 2. (I am not sure about this one)

III:-

Simplifying, sqrt(x)-sqrt(y)>sqrt(x+y)

Now, x and y are positive. Therefore, rHS >0 => Sqrt (x) > sqrt(y)

Also, squaring both the sides, x+y-2sqrt(xy) > x+y => -2sqrt(xy)>0 ====> Impossible for any value of x and y because x and y are positive.

Is this method correct? Please let me know.

[Disclaimer: I wouldn't use this method on the GMAT. This is just for learning.]

May be it's helpful for you