If u(u+v) different from 0 and u >0, is 1/(u+v) < 1/u + v?

[ghong14]

If u(u+v) is not equal to 0 and u >0, is 1/(u+v) < 1/u + v?

1) u+v >0

2) v>0

OA: B

My questions is even if we know that V > 0 and U>0 How would we know that 1/(u+v) < 1/u + v?

I simplified the later to 1/ (u+v) < (1 + UV) /u. Which comes out to be U < (1+vu)(U+V). IS this the right approach?

[jlucero]

If u(u+v) is not equal to 0 and u >0, is 1/(u+v) < 1/u + v?

1) u+v >0

2) v>0

OA: B

My questions is even if we know that V > 0 and U>0 How would we know that 1/(u+v) < 1/u + v?

I simplified the later to 1/ (u+v) < (1 + UV) /u. Which comes out to be U < (1+vu)(U+V). IS this the right approach?

It looks like your miscalculation is here:
1/u + v = v/uv (not (1 + v)/uv) + u/uv = (v + u) / uv

[ghong14]

It looks like your miscalculation is here:
1/u + v = v/uv (not (1 + v)/uv) + u/uv = (v + u) / uv

Hi I think that is still not right

1/U+V= (1+UV)/U

[RonPurewal]

If u(u+v) is not equal to 0 and u >0, is 1/(u+v) < 1/u + v?

The issue here seems to be the interpretation of the purple thing, above.
I'm going to follow order of operations and assume it means
(1/u) + v.

Still, when you post mathematical expressions on a forum, please use extra parentheses, when appropriate, to help clarify notation.

My questions is even if we know that V > 0 and U>0 How would we know that 1/(u+v) < 1/u + v?

Ok, so I'm lazy and I don't like doing algebra. (In Ron's world, even making the common denominator here is "a lot of work".)

So here's what I did with this one:
* I looked at the right-hand expression (the purple expression) first, and thought, "ok, well, that's definitely more than 1/u." (i also realized that it was more than v, but there's no v in the numerator of the other expression.)
* Then I looked at the left-hand expression and realized that it's less than 1/u, since the denominator is more than u.

Done.