Can someone help me to deduce this?
For any numbers x,y,z:
UPDATE: problem is wrong see coment.
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Codeforces (c) Copyright 2010-2026 Михаил Мирзаянов
Соревнования по программированию 2.0
Время на сервере: 19.04.2026 20:43:43 (g2).
Десктопная версия, переключиться на мобильную.
При поддержке
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 :)