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 | Benq | 3676 |
| 2 | Kevin114514 | 3659 |
| 3 | jiangly | 3640 |
| 4 | turmax | 3450 |
| 5 | tourist | 3408 |
| 6 | strapple | 3360 |
| 7 | Um_nik | 3354 |
| 8 | squareOf105 | 3348 |
| 9 | maroonrk | 3312 |
| 10 | idontwannadothis…e | 3191 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | Qingyu | 155 |
| 2 | nik_exists | 151 |
| 3 | maspy | 149 |
| 4 | AmShZ | 145 |
| 5 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | BledDest | 136 |
| 8 | adamant | 135 |
| 9 | maroonrk | 133 |
| 10 | cry | 130 |
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 :)