1
$\begingroup$

As the title states, the question is: Prove that if a and b are positive real numbers, then $a + b \geq ab$

For this proof, I'm supposed to prove by contrapositive.

So, I get this as a general statement:

Assume $a + b < ab$, then we will prove that a and b are real numbers that are less than 0.

But I'm wondering where do I go from here? The only thing I can think of is if a is negative and b is negative, multiplying two negatives will be positive. But then how do I actually write a formal proof for this?

$\endgroup$
3
  • 2
    $\begingroup$ The assertion is obviously false. Any more assumptions? $\endgroup$
    – MooS
    Commented Feb 3, 2015 at 8:35
  • $\begingroup$ @MooS My assertion or the original questions assertion? $\endgroup$
    – Alex
    Commented Feb 3, 2015 at 8:36
  • $\begingroup$ A similar inequality which is true is the AM–GM inequality. $\endgroup$
    – lhf
    Commented Feb 3, 2015 at 9:52

2 Answers 2

5
$\begingroup$

This is false. Take $a=b=3$. Then $a+b=6<9=ab$.

$\endgroup$
0
$\begingroup$

$a$ and $b$ are given as positive in the statement of the question, so they can't be negative. To analyse the statement further you can note:

The statement is equivalent to $$1\ge (a-1)(b-1)$$ or, if $a\gt 1$ $$b\le \frac a{a-1}$$ while if $a\lt 1$ $$b\ge \frac a{a-1}$$

If $a=1$ the original inequality becomes $b+1\ge b$ which is clearly true.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .