-1
$\begingroup$

In our base math course we had this question: if $a,b > 1$ then $1+ab > a+b$? There was also a hint at the bottom of the question to look at this expression: $(1+ab)-(a+b)$

I had tried to prove it several times, but I have no idea how to prove it mathematically. Logically, if $a,b>1$ then $a+b>1$ and $ab>1$. I looked at the expression and noticed that for every a,b as long as $a \neq b$ then $ab > a+b$.

We are not sure though if $a \neq b$. But that doesn't matter in both cases because even if $a=b$ the proof we have to get is for $1+ab$ and not $ab$. So we can for sure tell that for any $a,b>1$ then $1+ab > a+b$

This is all fun and games, but it is too much to write in a mathematical proof. So I would be happy if someone could shorten it to a mathematical in-line proof.

  • 0
    Your argument if $a=b$ then $ab=a+b$ is incorrect. $ab=a+b$ for $a=b=2$. What about 3? 4? etc. – 2017-01-07
  • 0
    Oh right, thanks. Didn't notice that so I will fix it. – 2017-01-07

3 Answers 3

4

we have $$1+ab>a+b$$ this is equivalent to $$1-a+b(a-1)>0$$ $$1-a-b(1-a)>0$$ and this is equivalent to $$(1-a)(1-b)>0$$ which is true

  • 0
    Could you explain a bit more the second step? I didn't really understand why the two expressions are equal? You just changed the + to a - and they are still equal? – 2017-01-07
  • 0
    yes i have made a typo and now it is corrected – 2017-01-07
  • 0
    Ok now I understand! Thanks a lot! – 2017-01-07
  • 0
    ok i'm waiting for your next problem now! – 2017-01-07
2

Hint: Expand the expression $(a-1)(b-1)$

1

Notice that $1+ab>a+b\implies (1-a)(1-b)>0$

So, knowing that $a,b>1$ you can say that $$(1-a)(1-b)>0$$ $$\implies 1+ab-a-b>0\implies 1+ab>a+b$$