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# If ab >0, is (ab)^2 < (ab)^(1/2) ?

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Q51  V47
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Q51  V47
Re: If ab >0, is (ab)^2 < (ab)^(1/2) ? [#permalink]
Abhi077 wrote:
If $$ab >0$$, is $$(ab) ^2$$ < $$\sqrt{ab}$$

1.$$\frac{4}{a}$$ > $$7b$$
2. $$a-16 > -16$$

With the correction, it's much easier to analyze the question algebraically. As I explained above, the only way (ab)^2 < √(ab) is true is if 0 < ab < 1, so that's what we want to know: is ab < 1?

Using only Statement 1, we have two cases:

- if a > 0, then we can multiply both sides of the inequality by a without reversing the inequality. So we then have

4/a > 7b
4 > 7ab
4/7 > ab

So in this case, it is indeed true that ab < 1.

- if a < 0, then we can again multiply both sides of the inequality by a, but because we're multiplying by a negative number, we must reverse the inequality:

4/a > 7b
4 < 7ab
4/7 < ab

So when a (and also b of course, since ab > 0) is negative, then the value of ab can be anything greater than 4/7, and can indeed be larger than 1. It is easy enough to confirm this just by choosing numbers: we might have, say, that a = -4 and b = -100, for example, in which case 4/a > 7b is clearly true, since -1 > -700, and in this case ab is much larger than 1.

Statement 2 alone is clearly not sufficient, since it only tells us a > 0. But once we know that a > 0, that rules out the second case above in our analysis of Statement 1. So we know only the first case is possible, and that ab < 4/7, so the two statements together are sufficient, and the answer is C.
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Re: If ab >0, is (ab)^2 < (ab)^(1/2) ? [#permalink]
Abhi077 wrote:
If $$ab >0$$, is $$(ab) ^2$$ < $$\sqrt{ab}$$

1.$$\frac{4}{a}$$ > $$7b$$
2. $$a-16 > -16$$

So i would say, nice , Lets see, given is important here ab> 0

Lets manipulate $$(ab)^4$$ < ab

take common and it becomes => (ab) {$$(ab)^3$$ -1 } < 0,For representation purpose only (x) (y) < 0

Now for the above to be true, it can be that x > 0 & y< 0 or x < 0 & y > 0, So from given, only bold is valid

which means we have to only focus on (ab)^3 < 1 part

from 1) $$\frac{4}{a}$$ > $$7b$$

Can be true when a=-1, b=-1, this makes the question as No
Can be true when a=1, b=-1, this makes the question as Yes

From 2) a-16 > -16, this can only be true when a> 0, dont know anything about B

When we combine we can directly pinpoint on the condition that

a > 0 and b < 0

C
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Re: If ab >0, is (ab)^2 < (ab)^(1/2) ? [#permalink]
Abhi077 wrote:
If $$ab >0$$, is $$(ab) ^2<\sqrt{ab}$$

(1) $$\frac{4}{a}>7b$$
(2) $$a-16 > -16$$

$$ab>0…{{a,b}}=same.sign≠0$$
$$(ab) ^2<\sqrt{ab}…ab=x…x^2<\sqrt{x}$$?
rearrange: $$x^2<\sqrt{x}…x^4<x…x(x^3-1)<0$$?
if $$x<0$$ is $$(x^3-1)>0…x^3>1…x>1…ab>1$$; invalid, since $$x<0$$;
if $$x>0$$ is $$(x^3-1)<0…x^3<1…x<1…ab<1$$? find this.

(1) $$\frac{4}{a}>7b$$:
if $${{a,b}}=positive…\frac{4}{a}>7b…4/7>ab…ab<1$$
if $${{a,b}}=negative…\frac{4}{-a}>7•(-b)…4/7<ab…ab>1$$
(2) $$a-16 > -16$$: $$a>0…b>0$$ but we don't know if $$ab<1$$, insufi.
(1&2): $$a>0$$ then $${{a,b}}=positive…\frac{4}{a}>7b…4/7>ab…ab<1$$; sufic.