Combining a and b

[kouranjelika]

Hi guys,

Can't seem to find this question listed anywhere. The answer is WAY over my head, I would really appreciate a good explanation and a civil way to do these generally.

Ron, I watched your three Inequalities Workshops from '10&'12, it helped, but as some of these get progressively harder, some of the rules or concepts are difficult to apply (as here).

"
If ab ≠ 0 and a + b ≠ 0, is ?

(1) |a| + |b| = a + b

(2) a > b
"

Thanks,
Anjelika

Source: MGMAT CATs

[RonPurewal]

There's no question here. There's nothing at all between "Is" and the question mark. Oops, try again.

[RonPurewal]

I presume you tried to copy and paste directly from the practice test.
If the problem contains nothing but "normal" characters"”no exponents, no vertical fractions, basically nothing you couldn't do on an old typewriter"”then copy/paste will work fine.
If there's any special forrmatting, though, the test software will contain that part as a picture, not as actual math symbols. You can't copy/paste a picture into a text-based forum, so that's not going to work here.

You'll have to do your best to transcribe the problem with your own keyboard.

[kouranjelika]

Shoot! My bad!!

Ok, so here it goes:

If ab cannot equal 0 (product of a & b) and a + b cannot equal zero (sum).
Is 1/(a+b) < 1/a + 1/b?

Stat (1): |a| + |b| = a + b
Stat (2): a>b

So I posted this before I have spent literally the entire week exhausting this subject.
I think I can actually try to answer it now.

So first thing I would do is combine the right side of the inequality in the question stem to have a common denominator. Although I don't particularly know that this helps me much right away.

1/(a+b) < b/ab + a/ab?
1/(a+b) < (b+a)/ab?

Now I would go to the statements:

Stat (1): |a| + |b| = a + b
This looks super whack to me, but I think it's just a trick. This simply means a & b MUST be positive, otherwise this equation will not work. So now I know a>0 & b>0.
Cool, I bring this knowledge back to the inequality:
if two positive numbers are added, they will DEF result in a positive number. So we can multiply both sides of the inequality in question by (a+b), yielding:
1<(a+b)^2/ab
Since we know a+b cannot be zero and a & b must be positive, even if they are the minimum value 1, that would be 2^2/1 = 4. SO, yes, larger than 1. I suppose we are not given a constraint that a & b must be integers. I wonder if I set them to say .1 each, it would still hold. Ok, I would try that, HA it turns out to be 4 again. So looks like it would always indeed hold to be true.
Sufficient.

The way the question explanation presented it was quite clever too, they made both sides of the inequality have the same denominator (which I think is better because you still don't have to multiply both sides of the inequality by an unknown number (a+b), doing which makes me feel all jittery inside).
So they made it into this:
ab/[ab(a+b)] < (a^2 + 2ab + b^2)/[ab(a+b)]?
Which was totally over my head a few days ago, but now it looks like amazing proof of the fact that the right side will indeed be larger as it has an extra ab, b^2 and a^2 in the numerator.

Stat (2): a>b
I would jump back to my original question Stem and plug numbers. Although something tells me that the number magnitude won't actually matter in this case.
Picking numbers I keep in mind that neither variable can be zero, and that the sum of the two variables cannot be zero either.
Case I: a=2 b=1
1/3 < 1/2 + 1? Yes
Case II: a=2 b=-1
1/(2-1) < 1/2 + 1/-1
1 < -1/2 Hello Tiger. NO!
Insufficient.

I would really appreciate if you could quickly glance through my reasoning Ron, but generally I think I feel ok on this question now.
Grazie mille per tutto tuo attenzione.

[RonPurewal]

It's not the easiest thread to find, but it's there.

post98573.html#p98573
A different approach to statement 1 is given there. Check it out.

[kouranjelika]

Awesome! I did look for it, couldn't find it though.

Thanks!

[RonPurewal]

Sure.

This thread is now locked; if there are other questions, please post them at the link. Thanks.