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I am trying to prove that if $x > 1$ and $y > 1$, then $xy >1$.

I am thinking that we can use proof by contradiction.

So we can assume that $xy≤1$ and, $x>1$ and $y>1$.

I got stuck and don't know what should be the next statement.

Any comment and answer you can provide would be greatly appreciated.

AYA
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    For such elementary proof, you have to be more precise about what is known. Otherwise I can just write : $x>1 \implies x \times y > 1\times y = y >1$. – nicomezi Jul 12 '20 at 10:10
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    Take $x=1+a,y=1+b$ . Then ? – user-492177 Jul 12 '20 at 10:10
  • @user710290 A very good approach. We have $a,b>0$ immediately giving the desired result. – Peter Jul 12 '20 at 10:16
  • @user710290, Thank you for your comment. Based on your suggestion, please let me know if this is correct. Since $x>1$ and $y>1$, then we can express $x=1+a$ and $y=1+b$ such that x and y are elements of the set of real numbers greater than 0. Hence, ... – AYA Jul 12 '20 at 10:23
  • To @user710290,

    $xy=(1+a)(1+b)$ $=1+x+y+xy$ Since $1+x+y+xy>1$, thus $xy>1$ QED.

     – AYA Jul 12 '20 at 10:26
  • @MAK Those should be $a$ and $b$ on right side of $=$ – user-492177 Jul 12 '20 at 10:33
  • Noted. Thank you @user710290 – AYA Jul 12 '20 at 10:38

3 Answers3

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$$(y-1)(x-1)>0 \Rightarrow yx> x+y-1>x>1$$

zkutch
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Let's take $x=t+1$ and $y=s+1$(here $t$ and $s$ are positive i.e >0)

we need to prove that $xy>1$

so $xy=(t+1)(s+1)$

$xy=st+s+t+1$

as we know $s$ and $t$ are positive then $s+t+st$ must be positive

Therefore,$xy>1$

Anonymous
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Because $$xy-1=xy-x+x-1=x(y-1)+(x-1)>0.$$

Michael Rozenberg
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