### LashaBukhnikashvili's blog

By LashaBukhnikashvili, 8 years ago, Can someone help me to deduce this?

For any numbers x,y,z: UPDATE: problem is wrong see coment. math, Comments (7)
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•  » » I thinking about this but,this is when x,y,z are positive other way no :(
•  » » » 8 years ago, # ^ | ← Rev. 2 →   When numbers can be negative, it's incorrect inequality. For example: x = 0,  y = 2,  z =  - 1
•  » » » » 8 years ago, # ^ | ← Rev. 3 →   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<0now I guess that this way is difficult. can somone tell me another easy way to prove first Inequality? •  » » » » » 8 years ago, # ^ | ← Rev. 3 →   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 :)