Can someone help me to deduce this?
For any numbers x,y,z:
UPDATE: problem is wrong see coment.
# | User | Rating |
---|---|---|
1 | tourist | 3985 |
2 | jiangly | 3741 |
3 | jqdai0815 | 3682 |
4 | Benq | 3529 |
5 | orzdevinwang | 3526 |
6 | ksun48 | 3489 |
7 | Radewoosh | 3483 |
8 | Kevin114514 | 3442 |
9 | ecnerwala | 3392 |
9 | Um_nik | 3392 |
# | User | Contrib. |
---|---|---|
1 | cry | 169 |
2 | maomao90 | 162 |
2 | Um_nik | 162 |
2 | atcoder_official | 162 |
5 | djm03178 | 158 |
6 | -is-this-fft- | 157 |
7 | adamant | 155 |
8 | awoo | 154 |
8 | Dominater069 | 154 |
10 | nor | 150 |
Can someone help me to deduce this?
For any numbers x,y,z:
UPDATE: problem is wrong see coment.
Name |
---|
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 :)