vittalgmat wrote: If x is positive, which of the following could be the correct ordering of 1/x,2x, and x^2

I. x^2<2x<1/x
II. x^2<1/x<2x
III. 2x< x^2<1/x

If you can quickly find a value of x that works in one of the inequalities, that may be the fastest way to establish that the inequality can be true. For II in particular it can be difficult to find a suitable number. If you can't find numbers that work, you can do this question algebraically; since x is positive, we can safely multiply or divide both sides of each inequality by x. Looking at I, we have three inequalities:

x^2 < 2x, or x < 2
2x < 1/x, or x^2 < 1/2, so x < sqrt(1/2)
x^2 < 1/x, or x^3 < 1, so x < 1

and if all three of these are true, I will be true - so I will be true if 0< x < sqrt(1/2)

For II, we have

x^2 < 1/x, so x^3 < 1, or x < 1
1/x < 2x, so 1 < 2x^2 or x > sqrt(1/2)
x^2 < 2x, so x < 2

and if all three of these are true, II will be true - so II will be true if sqrt(1/2) < x < 1

For III we have

2x < x^2, so 2 < x
x^2 < 1/x, so x^3 < 1, or x < 1

and since x cannot be larger than 2 and less than 1 at the same time, III cannot possibly be true.

nice, ian.

re: these two approaches:

Ian Stewart wrote:x^2 < 2x, or x < 2
2x < 1/x, or x^2 < 1/2, so x < sqrt(1/2)
x^2 < 1/x, or x^3 < 1, so x < 1

and

x^2 < 1/x, so x^3 < 1, or x < 1
1/x < 2x, so 1 < 2x^2 or x > sqrt(1/2)
x^2 < 2x, so x < 2

if you have a "sandwich inequality" of the type A < B < C, you don't have to check all three of these inequalities.
it's sufficient to check just "A < B" and "B < C".
any number that satisfies both of these will automatically satisfy A < C, since B bigger than A and C bigger than B implies that C is bigger than A.

so these approaches would be just as effective if you ignored the 3rd inequality of each one.

aside from that, though -- not much else you could improve (i.e., not really a shorter method) unless you are insanely good at picking numbers.