p, q, r and s are any four positive real numbers, the minimum value of

Question

p, q, r and s are any four positive real numbers, the minimum value of  \frac{p}{q} + \frac{q}{r} + \frac{r}{s} + \frac{s}{p}  is

(A) 0 
(B) 1 
(C) 2 
(D) 2√2
(E) 4

IanStewart · Answer

I'd be completely shocked if this problem could be solved with only GMAT level math. It's related to a much simpler problem, which can be solved using GMAT level math, and anyone who has worked through the question below will probably be able to correctly guess the answer to the question above: If x > 0, what is the minimum possible value of x + \frac{1}{x} ? There are at least three ways to solve this, one using GMAT-level math, and two using core techniques from math beyond GMAT level: 1. Clearly x + (1/x) can equal 2, when x = 1. If it could equal something less than 2, then this inequality would be true: x + \frac{1}{x} < 2 but if we multiply on both sides by x (which is fine, since x > 0) and rearrange to get a quadratic, this leads us to \begin{align} x + \frac{1}{x} &< 2 \ x^2 + 1 &< 2x \ x^2 - 2x + 1 &< 0 \ (x - 1)^2 &< 0 \end{align} which is clearly impossible -- a square can't be negative. So the minimum value of x + (1/x) cannot be less than 2, and since it can equal 2, then 2 is its minimum value. 2. If you know calculus, you can just take the derivative of f(x) = x + (1/x), and set the derivative equal to zero, to find the value of x that minimizes f(x) (you'd get two solutions, 1 and -1, but x > 0 here, so the minimum happens when x = 1, and the minimum is therefore 2). 3. One of the most famous results in more advanced statistics is the Arithmetic Mean-Geometric Mean inequality (the "AM-GM inequality"). That inequality just says that the arithmetic mean (the normal average) of a set of positive numbers is always greater than or equal to the "geometric mean" of those numbers. If you have a set of n numbers, their "geometric mean" is just the nth root of their product. So the geometric mean of 11, 12 and 13 would be the 3rd root, or cube root, of 11*12*13, for example, which is less than their ordinary average, which is 12. It's easy to see, if you take any two positive numbers, that this inequality is true -- if you take, say, 3 and 5, the arithmetic mean is 4, while the geometric mean is the 2nd root, or square root, of 3*5 = 15, and since √15 < 4, the geometric mean is smaller than the arithmetic mean, as expected. The AM and GM are only equal when all your numbers are the same. If you just apply the AM-GM inequality to x and 1/x, their normal average is greater than or equal to their geometric mean. But their geometric mean is the square root of their product, and their product is 1. So their geometric mean is 1, and the average of x and 1/x must be greater than or equal to 1. So the sum of x and 1/x is greater than or equal to 2. That's a bit of a long preamble, but anyone who has thought through the AM-GM inequality solution to the simpler question above will probably see how to solve the question here -- you just do the same thing. If you take the four numbers p/q, q/r, r/s and s/p, their product is 1. Their geometric mean is the 4th root of their product, so is also 1. The arithmetic mean of the four fractions is greater than or equal to their geometric mean, so is greater than or equal to 1. Since we have 4 terms, that means their sum, which is precisely what we're trying to minimize, is greater than or equal to 4. Since their sum can in fact equal 4, when p = q = r = s, that's the minimum value. It looks to me like you can also solve using calculus - if you take partial derivatives with respect to p and with respect to q to locate the minimum, then plug in to compute the minimum, I think you end up with something like s + 1/s + r + 1/r, which must be (as proved in the simpler problem above) at least 2 + 2 = 4. There might be some technical issues with this approach though - I haven't done that kind of math in a long time, so it's very possible I've forgotten some important details about it. Anyway, these techniques are not just slightly out of GMAT scope, they're miles beyond the GMAT. The simpler problem I included at the start of this post is, however, possibly a useful question to look at.

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p, q, r and s are any four positive real numbers, the minimum value of

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p, q, r and s are any four positive real numbers, the minimum value of

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