Problem Solving

How to approach these kind of problems?

substituting numbers for x and y helps me.

You should consider it is true and start the different inequalities: Z > W
and then try to work on it to see whether it is true or not. Use squares, cross-multiplying...

pepeprepa wrote:You should consider it is true and start the different inequalities: Z > W
and then try to work on it to see whether it is true or not. Use squares, cross-multiplying...

..thanks..that looks like a good idea though it will be time consuming as we have to evaluate 3 eqns..thats like solving 3 problems in 1 problem time...anyway something is better than nothing...thanks!!

Hi pepeprepa
Intersting approach ! i would be glad to see if u could solve the very problem mentioned !!

I would take it as an example

Vishu

bha, I tam sorry if my way does not satisfy you! If I look at the problem some seconds and I have a method which can enable me to solve it in less than 2 minutes I do it. Unfortunately, we cannot always have a better choice than what we think during the first seconds. But if you can have the right answer and not lose time that's enough.

I.
1/sqrt(x+y) < sqrt(x+y)/2x
2x<x+y
x<y
It can be wrong, don't keep it.

II.
1/sqrt(x+y) < (sqrt(x) + sqrt(y))/(x+y)
sqrt(x+y) < sqrt(x) + sqrt(y)
x+y < x + 2*sqrt(y)*sqrt(x) + y
That must be true

III.
1/sqrt(x+y) < (sqrt(x) - sqrt(y))/(x+y)
sqrt(x+y) < sqrt(x) - sqrt(y)
x+y < x - 2*sqrt(y)*sqrt(x) + y
That's wrong, don't keep it

pepeprepa wrote:bha, I tam sorry if my way does not satisfy you! If I look at the problem some seconds and I have a method which can enable me to solve it in less than 2 minutes I do it. Unfortunately, we cannot always have a better choice than what we think during the first seconds. But if you can have the right answer and not lose time that's enough.

I.
1/sqrt(x+y) < sqrt(x+y)/2x
2x<x+y
x<y
It can be wrong, don't keep it.

II.
1/sqrt(x+y) < (sqrt(x) + sqrt(y))/(x+y)
sqrt(x+y) < sqrt(x) + sqrt(y)
x+y < x + 2*sqrt(y)*sqrt(x) + y
That must be true

III.
1/sqrt(x+y) < (sqrt(x) - sqrt(y))/(x+y)
sqrt(x+y) < sqrt(x) - sqrt(y)
x+y < x - 2*sqrt(y)*sqrt(x) + y
That's wrong, don't keep it

For number II and III can you say how you went from step 1 to step 2? For soe reason I am not seeing that but definitely see step 2 to step 3.

Thanks!

pepeprepa wrote: II.
1/sqrt(x+y) < (sqrt(x) + sqrt(y))/(x+y)
sqrt(x+y) < sqrt(x) + sqrt(y)
x+y < x + 2*sqrt(y)*sqrt(x) + y

Could you please dicribe this reasoning step-by-step

Thanx!

I solve this question by using 16 for x and 9 for y

II.
1/sqrt(x+y) < (sqrt(x) + sqrt(y))/(x+y)
I multiply both sides by something positive: (x+y)
sqrt(x+y) < sqrt(x) + sqrt(y)
I put on/at ?? square both sides (they are positive)
x+y < x + 2*sqrt(y)*sqrt(x) + y

pepeprepa wrote:bha, I tam sorry if my way does not satisfy you! If I look at the problem some seconds and I have a method which can enable me to solve it in less than 2 minutes I do it. Unfortunately, we cannot always have a better choice than what we think during the first seconds. But if you can have the right answer and not lose time that's enough.

No complaints mate...

Wow tough to see that one in less than 2 mins. BHA - out of curiosity ... how were you doing prior to this question. Trying to gauge the difficulty.

in less than a minute. you should evaluate all 4 ecuations.

pretend that x=1 and y=1.

the result of the question is 1/2

now the issue is to find something bigger than 1/2

the first one gives you 1/2 so is no bigger.
the second one gives you 2,8/4 so is bigger.
the trird one gives to you o/something, so is not bigger.

the logical answer is the second one.