mmcooley33 wrote:

walker wrote:

Let's consider original statement: \(\frac{1}{\sqrt{x+y}}\)

How can we approach the problem fast? Let's see when the original statement is very large : x,y ---> 0 and \(\frac{1}{\sqrt{x+y}}\) goes to infinity.

Now, let's see what do our options at x,y ---> 0.

I) \(\frac{\sqrt{x+y}}{2}\) goes to 0 at x,y ---> 0.

II) \(\frac{\sqrt{x}+\sqrt{y}}{2}\) goes to 0 at x,y ---> 0.

III) \(\frac{\sqrt{x}-sqrt{y}}{x+y}\) hm... it has x+y as a denominator. But what if x=y? At x=y it equals 0.

So, none of the options.

Can you explain how you used this in more detail, and perhaps give me some idea of how to apply this same principle going forward. Thanks a lot, i appreciate the help.

Just to add couple of words to Walker's great solution:

Note that we are asked "which of the following MUST be greater than \(\frac{1}{\sqrt{x+y}}\)?" not COULD be greater.

"MUST BE TRUE" questions:

These questions ask which of the following

MUST be true (must be greater in our case), or which of the following is

ALWAYS true no matter what set of numbers you choose. Generally for such kind of questions

if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

As for "COULD BE TRUE" questions:

The questions asking which of the following

COULD be true are different:

if you can prove that a statement is true for one particular set of numbers, it will mean that this statement could be true and hence is a correct answer.

So, if we find even one set of \(x\) and \(y\) for which \(\frac{1}{\sqrt{x+y}}\) is greater than option I for example then it'll mean that option I is not ALWAYS greater then \(\frac{1}{\sqrt{x+y}}\).

How can we increase the value of \(\frac{1}{\sqrt{x+y}}\)? Testing extreme examples: if \(x\) and \(y\) are very small (when their values approach zero) then denominator approaches zero and thus the value of the fraction goes to +infinity, becomes

very large. In this case:

I. \(\frac{\sqrt{x+y}}{2}\): nominator approaches zero, so the value of the whole fraction approaches zero as well, becoming

very small. So this option is not always more than given fraction;

II. \(\frac{\sqrt{x}+\sqrt{y}}{2}\): the same here - nominator approaches zero, so the value of the whole fraction approaches zero as well, becoming

very small. So this option is not always more than given fraction;

As for:

III. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\), if \(x=y\) then this fraction equals to zero and \(\frac{1}{\sqrt{x+y}}\) has some value more than zero, so this option also is not always more than given fraction;

Answer: none of the options must be greater than the given fraction.

Hope it's clear.

. Thanks for pointing out use of extreme cases for ruling out options.