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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2did you know that $(rk)/(r+k-1) > (rk)/(r+1-1) = k > 1$ by assumption – Christian Chapman May 15 '17 at 09:06
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1$$k>1 \rightarrow r+k-1>r+1-1=r \rightarrow \frac{rk}{r+k-1}< \frac{rk}{r}=k$$ So, your idea doesn't work. – 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 $$ \frac{rk}{r+k-1} = \frac{(a+1)(b+1)}{a+b+1} = \frac{ab +a+b+1}{a+b+1} > 1. $$
Rigel
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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 $k,r>1$
blue
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Would be nifty if you should the factoring prior to the last step. – Daniel R. Collins May 14 '17 at 14:21
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$0 \lt \cfrac{1}{r} \lt 1\,$ and $0 \lt \cfrac{1}{k} \lt 1\,$ since $r,k > 1$, therefore:
$$\frac{rk}{r+k-1} = \frac{1}{\cfrac{1}{r}+\cfrac{1}{k}-\cfrac{1}{rk}} = \frac{1}{1 - \left(1-\cfrac{1}{r}\right)\left(1-\cfrac{1}{k}\right)} > \frac{1}{1} = 1$$
dxiv
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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.
Laurent Duval
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