## View Poll Results: how are you at math?

Voters
88. You may not vote on this poll
• I am NFP: talented

14 15.91%
• I am NFP: above average

14 15.91%
• I am NFP: average

7 7.95%
• I am NFP: below average

2 2.27%
• I am NFP: I suck at math

12 13.64%
• I am NFJ: talented

10 11.36%
• I am NFJ: above average

15 17.05%
• I am NFJ: average

9 10.23%
• I am NFJ: below average

3 3.41%
• I am NFJ: I suck at math!

2 2.27%

# Thread: Any NFs Talented at Math?

1. I seriously suck at Maths! It's always been my weakest subject. My strongest subjects were English, Drama, Child Studies.

2. Oh God, I suck at math.

But I'm great at Sudoku.

3. I can learn certain math subjects quickly if I'm forced to learn it. Otherwise it's booooring. Not really my subject.

4. I loved it, calculus included. English was my minor in college though.

5. Aha, I found the problem, I think, that you mentioned on my visitor board.

Originally Posted by Chloee

prove that for ANY (real) positive numbers a,b,c and any not-negative number p this is true:

Here is my solution:

6. wow great!
i will read it later, but I already see you did it the sa way i did - with grouping in beginning and then switching a*p,b*p,c*p, with b*p.
but with this you prove it's only true in cases when a,b,c is bigger than 1 ?

7. ENFPs do have the capacity to do well with math. That's not going to transform me into one. When I think ENFP, I think dog. I like dogs. Wouldn't want to be one.

8. Originally Posted by Chloee
wow great!
i will read it later, but I already see you did it the sa way i did - with grouping in beginning and then switching a*p,b*p,c*p, with b*p.
but with this you prove it's only true in cases when a,b,c is bigger than 1 ?
I don't think the proof suffers from that limitation. I believe it works for all positive a,b,and c.

The trick is to impose on order on a,b, and c without loss of generality. At that point, the inequality is true.

After that using the fact that the sum of squares of real numbers has to be non-negative provides the rest of what is needed.

9. Originally Posted by ygolo
Aha, I found the problem, I think, that you mentioned on my visitor board.

Here is my solution:

My brain just exploded...excuse me while I clean it up...

10. Originally Posted by ygolo
I don't think the proof suffers from that limitation. I believe it works for all positive a,b,and c.

The trick is to impose on order on a,b, and c without loss of generality. At that point, the inequality is true.

After that using the fact that the sum of squares of real numbers has to be non-negative provides the rest of what is needed.
yep, i read it now, it's true.
that prove was 2003 in National competition in math here in Croatia for 1st year High School. i did it the same as you did until
1 b*p[(a2 -bc) + (b2-ac) + (c2-ab)] >= 0
then i've put
2 (a-b)*2= a2-2ab+b2 >= 0
3 (b-c)*2= b2-2bc+c2 >= 0
4 (c-b)*2= c2-2bc+b2 >= 0
since 2+3+4
is (2a2-2bc)+(2b2-2ac)+(2c2-2ab)>=0 which makes 1 true, after you divide it with 2 etc.

i like it because you dont have to have any real knowledge to do it, besides basic calculus. inequalities were always most interesting thing in math to me.

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