I seriously suck at Maths! It's always been my weakest subject. My strongest subjects were English, Drama, Child Studies.
I am NFP: talented
I am NFP: above average
I am NFP: average
I am NFP: below average
I am NFP: I suck at math
I am NFJ: talented
I am NFJ: above average
I am NFJ: average
I am NFJ: below average
I am NFJ: I suck at math!
I seriously suck at Maths! It's always been my weakest subject. My strongest subjects were English, Drama, Child Studies.
I can learn certain math subjects quickly if I'm forced to learn it. Otherwise it's booooring. Not really my subject.
I loved it, calculus included. English was my minor in college though.
thinking of you
Accept the past. Live for the present. Look forward to the future.
Robot Fusion
"As our island of knowledge grows, so does the shore of our ignorance." John Wheeler
"[A] scientist looking at nonscientific problems is just as dumb as the next guy." Richard Feynman
"[P]etabytes of [] data is not the same thing as understanding emergent mechanisms and structures." Jim Crutchfield
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 ?
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.
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.
Accept the past. Live for the present. Look forward to the future.
Robot Fusion
"As our island of knowledge grows, so does the shore of our ignorance." John Wheeler
"[A] scientist looking at nonscientific problems is just as dumb as the next guy." Richard Feynman
"[P]etabytes of [] data is not the same thing as understanding emergent mechanisms and structures." Jim Crutchfield
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.