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If $a < b$, which of the following are always correct?

1) $a^3< a^2b$

The answer to this question would be true because there is theorem that "If $a < b$ and we add the same positive number to both sides of the inequalitie, this would not effect the inequality."

Now the second problem is interesting:

2) $a^2 < ab$

Now there is also another theorem that says "If $a < b$ and we multiply the same positive number on both sides this would not effect the inequality."

Now, as it is specified on the top of the question:$a < b$. now in this question we are multiplying a on both sides which makes it $a^2 < ab$ so this should be correct. But, in the answers at the end of the book, the question number of this question is not given. Please help.

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Please re-title this question. Something along the lines of "solving simple inequality in two variables" would be more appropriate. – Tom Stephens Aug 11 '10 at 14:01
up vote 3 down vote accepted

I think the easiest way to see this is to bring everything to the left side. For inequality 1 you get:

a^3 - a^2 b < 0

Now factor out a^2.

a^2( a - b) < 0

This is a true statement - provided a != 0 - since a^2 is positive and a-b is negative. Recall that a < b.

For inequality 2 you get:

a (a - b) < 0

While (a - b) is negative we can't say whether or not a is positive or negative. The inequality is true only if a is positive and is false if a is negative.

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exelent answer, i understood the concept now, that how to solve such problems. thank you very much – Zia ur Rahman Aug 11 '10 at 20:15

HINT If $b-a > 0 $ then $c(b-a) > 0 \iff c > 0$. Now specialize $c = a, \; a^2$.

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1) a^3< a^2b

is always true because even if 'a' is negative

a^3 will always be less than a^2b (positive)

In 2, if you have for example

a = -10

b = 2

then a < b = true

a^2 = 20

ab = -20

then a^2 < ab becomes false

so 1 is always true, 2 is not.

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$a^3< a^2b$ is false for $a=0$. – Rasmus Aug 11 '10 at 15:33

Both are false; try for instance for a = 0 < 1 = b.

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