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I am given with real numbers $r,k > 1$. Is the following true? $$\frac{rk}{r+k-1} > 1$$ I have taken a few reals and the result holds for them. Does this inequality holds in general or not?

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    $\begingroup$ did you know that $(rk)/(r+k-1) > (rk)/(r+1-1) = k > 1$ by assumption $\endgroup$ – enthdegree May 15 '17 at 9:06
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    $\begingroup$ $$k>1 \rightarrow r+k-1>r+1-1=r \rightarrow \frac{rk}{r+k-1}< \frac{rk}{r}=k$$ So, your idea doesn't work. $\endgroup$ – blue May 15 '17 at 12:00
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The inequality is trivial if you write $a := r-1>0$, $b := k-1>0$. Namely $$ \frac{rk}{r+k-1} = \frac{(a+1)(b+1)}{a+b+1} = \frac{ab +a+b+1}{a+b+1} > 1. $$

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$$\frac{rk}{r+k-1}>1 \leftrightarrow \\ rk>r + k -1 \leftrightarrow \\ rk -r>k-1 \leftrightarrow \\ r(k-1)>k-1\leftrightarrow\\ r>1$$ These equations are equivalent. Note that the reason we can say this is: $r+k-1>0$ and $k-1>0$. So yes, it holds for every $k,r>1$

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  • $\begingroup$ Would be nifty if you should the factoring prior to the last step. $\endgroup$ – Daniel R. Collins May 14 '17 at 14:21
  • $\begingroup$ @DanielR.Collins It's simple. You are just taking an $r$ out. $\endgroup$ – Obinna Nwakwue May 14 '17 at 19:30
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$0 \lt \cfrac{1}{r} \lt 1\,$ and $0 \lt \cfrac{1}{k} \lt 1\,$ since $r,k > 1$, therefore:

$$\frac{rk}{r+k-1} = \frac{1}{\cfrac{1}{r}+\cfrac{1}{k}-\cfrac{1}{rk}} = \frac{1}{1 - \left(1-\cfrac{1}{r}\right)\left(1-\cfrac{1}{k}\right)} > \frac{1}{1} = 1$$

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You can first use the $r,k>1$ hypothesis wisely, by shifting to $0$. For instance with $r=r'+1$ and $k=k'+1$ (and $r',k'>0$), since $r+k-1> 0$, you end up checking if:

$$ rk-r-k+1 >0$$

or

$$ (r'+1)(k'+1)-r'-k'-1 >0$$

and simplifying:

$$ r'k'>0$$

which is the hypothesis.

Now you have found the path, you can go backward. So by hypothesis:

$$(r-1)(k-1)>0 $$

then

$$rk-r-k+1>0 $$

thus

$$rk>-(-r-k+1) $$

and since $r+k-1> 0 $, you get your result directly.

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