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Is 1/p > r/(r^2 + 2)?

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Joined: 18 Feb 2015
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Is 1/p > r/(r^2 + 2)? [#permalink]

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27 Nov 2016, 12:01
Hi

When multiplying the following inequality 1/r > r/r^2+2 We get R^2 +2 > R^2. On the next step why do we divide R^2 +2 by R^2 ? As I have learned from other problems, since we do not know the sign of R, we should never multiply it on the other side and make the whole equation greater than or equal to 0 etc. So as per that, I thought that the last equation (R^2 +2 by R^2) would be R^2 +2 -R^2 >0 (So instead of dividing,I subtracted it). What am I missing?

Thanks

Bunuel wrote:
This question was posted in another thread just an hour ago, here is my reply to that post.

Is $$\frac{1}{p}>\frac{r}{r^2 + 2}$$?

(1) $$p=r$$ --> $$\frac{1}{r}>\frac{r}{r^2+2}?$$ --> as $$r^2+2$$ is always positive, multiplying inequality by this expression we'll get: $$\frac{r^2+2}{r}>r?$$ --> $$r+\frac{2}{r}>r?$$ --> $$\frac{2}{r}>0?$$. This inequality is true when $$r>0$$ and not true when $$r<0$$. Not sufficient.

(2) $$r>0$$. Not sufficient by itself.

(1)+(2) $$r>0$$, $$\frac{2}{r}>0$$. Sufficient.

tejal777 again when you simplifying $$\frac{1}{p} >\frac{r}{r^2 +2}$$ to $$r^2 +2 > pr$$, you are making a mistake. You can multiply inequality by r^2 +2 as this expression is always positive, BUT you can not multiply inequality by p, as we don't know the sigh of p.

Hope it's clear.
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Joined: 02 Sep 2009
Posts: 44351
Re: Is 1/p > r/(r^2 + 2)? [#permalink]

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28 Nov 2016, 01:37
HarveyKlaus wrote:
Hi

When multiplying the following inequality 1/r > r/r^2+2 We get R^2 +2 > R^2. On the next step why do we divide R^2 +2 by R^2 ? As I have learned from other problems, since we do not know the sign of R, we should never multiply it on the other side and make the whole equation greater than or equal to 0 etc. So as per that, I thought that the last equation (R^2 +2 by R^2) would be R^2 +2 -R^2 >0 (So instead of dividing,I subtracted it). What am I missing?

Thanks

Bunuel wrote:
This question was posted in another thread just an hour ago, here is my reply to that post.

Is $$\frac{1}{p}>\frac{r}{r^2 + 2}$$?

(1) $$p=r$$ --> $$\frac{1}{r}>\frac{r}{r^2+2}?$$ --> as $$r^2+2$$ is always positive, multiplying inequality by this expression we'll get: $$\frac{r^2+2}{r}>r?$$ --> $$r+\frac{2}{r}>r?$$ --> $$\frac{2}{r}>0?$$. This inequality is true when $$r>0$$ and not true when $$r<0$$. Not sufficient.

(2) $$r>0$$. Not sufficient by itself.

(1)+(2) $$r>0$$, $$\frac{2}{r}>0$$. Sufficient.

tejal777 again when you simplifying $$\frac{1}{p} >\frac{r}{r^2 +2}$$ to $$r^2 +2 > pr$$, you are making a mistake. You can multiply inequality by r^2 +2 as this expression is always positive, BUT you can not multiply inequality by p, as we don't know the sigh of p.

Hope it's clear.

We are not doing that. We simply write (r^2 + 2)/r as r + 2/r ((r^2 + 2)/r = r^2/r + 2/r = r + 2/r).
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Re: Is 1/p > r/(r^2 + 2)? [#permalink]

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05 Jan 2018, 02:51
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Is 1/p > r/(r^2 + 2)?   [#permalink] 05 Jan 2018, 02:51

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