On some question types, plugging in numbers can be a very good strategy. But on questions that ask 'what must be true', or 'what must be larger', it usually isn't an ideal approach. If you pick values for x and y here and plug them into each expression, you may find, say, that I and II are larger than the expression in the question. All you've shown is that I and II are sometimes larger. You still can't be sure that they must be larger. There's an infinite number of possible values for x and y, so how many numbers do you need to try to be sure of your answer?

That said, plugging in numbers is much better than guessing randomly- in most cases, you'll eliminate some wrong answers this way, and often enough, it will lead you to the right answer. 'Often enough' isn't good enough for me, so I'd rather use an approach that is guaranteed to get me the right answer all the time.

For this question, I first would look ahead at I, II and III, and compare them with the expression in the question. I, II and III have rational denominators- no square roots- while the expression in the question has a root in the denominator. Comparing these expressions will be a lot easier if the denominators are made more similar. So the first thing I'd do is rationalize the denominator in the expression in the question by multiplying the numerator and denominator by root(x+y):

1/root(x+y) = root(x+y)/(x+y) = Q

I'm going to call this 'Q' so I don't need to keep writing "the expression in the question"!

Now this looks a lot more like I, II and III. Let's look at each:

I: the only difference between I and Q is the denominator. If x>y, 2x (which is just x+x) will be larger than x+y, and otherwise it will be smaller or equal, so sometimes I will be smaller than Q, sometimes not. It is not true that I must be greater than Q.

II: Here, the denominators are the same; we just want to know if root(x) + root(y) is always larger than root(x+y). There are a few ways to see this. For example, multiply this inequality by root(x+y) on both sides:

Is root(x) + root(y) > root(x+y) ?
--> is root(x^2 + xy) + root(xy + y^2) > x + y ?

Well, root(x^2) = x, so root(x^2 + y) must be larger than x. Similarly, root(x+y^2) must be greater than y. So this inequality must be true.

Or, since we're dealing only with positive numbers, we can square both sides of the inequality:

Is [root(x) + root(y)]^2 > [root(x+y)]^2 ?
--> is x + 2root(xy) + y > x+y ?

And since x and y are positive, this must be true.


III: Notice that the numerator of III can easily be negative, while Q must be positive, so III will not always be greater than Q.

beeparoo wrote:

Ian Stewart wrote:II: Here, the denominators are the same; we just want to know if root(x) + root(y) is always larger than root(x+y). There are a few ways to see this. For example, multiply this inequality by root(x+y) on both sides:

Is root(x) + root(y) > root(x+y) ?
--> is root(x^2 + y) + root(x+y^2) > x + y ?

Hi Ian, I'm having trouble understanding how you derived the left-hand side of the inequality, namely: root(x^2 + y) + root(x+y^2)

Going through your approach, step-by-step, I reached this part and stopped short because I derived

root(x^2 + xy) + root(xy + y^2)

While I don't think it changes the outcome of your results, I'm really curious if I am missing something... Eep!

Nice catch, beeparoo! Thanks for spotting that. No, it doesn't change the result or logic in any way, but I'll edit the post to fix the typo.