ygolo
My termites win
- Joined
- Aug 6, 2007
- Messages
- 6,003
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.
Your way is simpler. I just typed out what popped into my head. Thought I think you made a typo on the fourth line.
Shouldn't it be 4 (c-a)^2= c^2-2ac+a^2 >= 0 ?