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

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Manager
Joined: 18 Feb 2015
Posts: 85
Is 1/p > r/(r^2 + 2)?  [#permalink]

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27 Nov 2016, 11:01
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: 51218
Re: Is 1/p > r/(r^2 + 2)?  [#permalink]

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28 Nov 2016, 00:37
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).
_________________
Manager
Joined: 10 Apr 2018
Posts: 180
Re: Is 1/p > r/(r^2 + 2)?  [#permalink]

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15 Sep 2018, 08:31
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
CEO
Joined: 11 Sep 2015
Posts: 3238
Re: Is 1/p > r/(r^2 + 2)?  [#permalink]

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09 Oct 2018, 05:58
Top Contributor
tejal777 wrote:
Is 1/p > r/(r² + 2) ?

(1) p = r
(2) r > 0

Target question: Is 1/p > r/(r² + 2) ?

Statement 1: p = r
Take the target question and replace p with r to get: Is 1/r > r/(r² + 2) ?
Since r² + 2 is always POSITIVE, we can safely multiply both side of the inequality by r² + 2.
When we do this, we get: Is (r² + 2)/r > r ?

Now we can apply the following fraction property: (a + b)/c = a/c + b/c
We get: Is r²/r + 2/r > r ?
Simplify to get: Is r + 2/r > r ?
Subtract r from both sides to get: Is 2/r > 0 ?
At this point, we can see that statement 1 is not sufficient.
When r is POSITIVE, the answer to our revised target question is YES, it IS the case that 2/r > 0
When r is NEGATIVE, the answer to our revised target question is NO, it is NOT the case that 2/r > 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: r > 0
Since we have no information about p, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
From statement 1 we learned that . . .
When r is POSITIVE, the answer to our revised target question is YES, it IS the case that 2/r > 0
When r is NEGATIVE, the answer to our revised target question is NO, it is NOT the case that 2/r > 0

Statement 2 tells us that r is POSITIVE
So, we can be certain that the answer to our revised target question is YES, it IS the case that 2/r > 0
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Cheers,
Brent
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Re: Is 1/p > r/(r^2 + 2)? &nbs [#permalink] 09 Oct 2018, 05:58

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

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