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

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Manager
Joined: 18 Feb 2015
Posts: 86
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.
Math Expert
Joined: 02 Sep 2009
Posts: 49208
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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Manager
Joined: 10 Apr 2018
Posts: 95
Re: Is 1/p > r/(r^2 + 2)?  [#permalink]

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15 Sep 2018, 09:31
Hi,

here are my two cents.

From Stmt 1:
p=r
Then
$$\frac{1}{r} > \frac{r}{r^2+2}$$

then

$$\frac{1}{r} - \frac{r}{r^2+2}$$>0

$$\frac{r^2+2 - r^2}{r(r^2+2)}$$ > 0

$$\frac{2}{r(r^2+2)}$$ > 0

So if we know that r > 0 then we can say that the exp is >0

But we don't know if r>0

Stmt 2:

r>0

That means $$\frac{r}{r^2+2}$$is positive or >0

Now question is IS $$\frac{1}{p}$$ >0 or p>0
Since we have no information about p hence insufficient.

Now combining Stmt 1 and Stm2 we have
p=r and r>o so p>0

Hence sufficient.

Probus
Re: Is 1/p > r/(r^2 + 2)? &nbs [#permalink] 15 Sep 2018, 09:31

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