Can someone help me to deduce this?
For any numbers x,y,z:
UPDATE: problem is wrong see coment.
Can someone help me to deduce this?
For any numbers x,y,z:
UPDATE: problem is wrong see coment.
№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 3690 |
2 | jiangly | 3647 |
3 | Benq | 3581 |
4 | orzdevinwang | 3570 |
5 | Geothermal | 3569 |
5 | cnnfls_csy | 3569 |
7 | Radewoosh | 3509 |
8 | ecnerwala | 3486 |
9 | jqdai0815 | 3474 |
10 | gyh20 | 3447 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | maomao90 | 174 |
2 | awoo | 164 |
3 | adamant | 163 |
4 | TheScrasse | 159 |
5 | nor | 157 |
6 | maroonrk | 155 |
7 | -is-this-fft- | 152 |
8 | Petr | 146 |
8 | orz | 146 |
10 | BledDest | 145 |
Название |
---|
http://en.wikipedia.org/wiki/Muirhead%27s_inequality
http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means
I thinking about this but,this is when x,y,z are positive other way no :(
When numbers can be negative, it's incorrect inequality. For example: x = 0, y = 2, z = - 1
you'r right,i had a mistake;
x,y,z is such numbers that in this numbers 2 of them is positive and one any number.
More formally at first i had such Inequality(where a,b,c>0):
than mentioned x=b+c-a,y=a+c-b,z=a+b-c numbers,and get:
may say that a<=b<=c and get that x>0 y>0 z<0
now I guess that this way is difficult. can somone tell me another easy way to prove first Inequality?
This is a well-known inequality, http://en.wikipedia.org/wiki/Schur's_inequality
It is equivalent to the following: (a + b - c)(b + c - a)(a + c - b) ≤ 3abc, so when one of the brackets is negative, left part of inequality is also negative, when the right part is always positive.
thanksss bro :)