Theorem: Schur's Inequality

Let $x, y, z \in \R_{\ge 0}$ be positive real numbers.
Let $t \in \R, t > 0$ be a (strictly) positive real number.
Then:
:$x^t \paren {x - y} \paren {x - z} + y^t \paren {y - z} \paren {y - x} + z^t \paren {z - x} \paren {z - y} \ge 0$
The equality holds {{iff}} either:
: $x = y = z$
: Two of them are equal and the other is zero.
When $t$ is a positive even integer, the inequality holds for ''all'' real numbers $x, y, z$.

Proof:
We note that the inequality, as stated, is symmetrical in $x, y$ and $z$.
{{WLOG}}, we can assume that $x \ge y \ge z \ge 0$.
Consider the expression:
:$\paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z}} + z^t \paren {x - z} \paren {y - z}$
We see that every term in the above is non-negative. So, directly:
:$(1): \quad \paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z}} + z^t \paren {x - z} \paren {y - z} \ge 0$
If $x = y = z$, all of $x - y$, $x - z$ and $y - z$ are $0$.
Thus equality holds.
Inspection on a case-by-case basis provides evidence for the other conditions for equality.
To show these are the only cases, we suppose $x, y, z$ are not equal.
Then $x > y > z \ge 0$.
We thus have $x - z > y - z$.
Hence:
{{begin-eqn}}
{{eqn | o = 
      | r = \paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z} } + z^t \paren {x - z} \paren {y - z}
}}
{{eqn | o = \ge
      | r = \paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z} }
      | c = second term is non-negative
}}
{{eqn | o = >
      | r = \paren {x - y} \paren {x^t \paren {y - z} - y^t \paren {y - z} }
      | c = 
}}
{{eqn | r = \paren {x - y} \paren {y - z} \paren {x^t  - y^t}
      | c = 
}}
{{eqn | o = >
      | r = 0
      | c = $x > y > z$
}}
{{end-eqn}}
Now we suppose two numbers are equal, but the other is neither the same number or $0$.
If $x = y > z > 0$:
{{begin-eqn}}
{{eqn | o = 
      | r = \paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z} } + z^t \paren {x - z} \paren {y - z}
}}
{{eqn | r = z^t \paren {x - z} \paren {y - z}
      | c = $x - y = 0$
}}
{{eqn | o = >
      | r = 0
      | c = $z > 0$, $x - z = y - z > 0$
}}
{{end-eqn}}
If $x > y = z \ge 0$:
{{begin-eqn}}
{{eqn | o = 
      | r = \paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z} } + z^t \paren {x - z} \paren {y - z}
}}
{{eqn | r = \paren {x - y} \paren {x^t \paren {x - z} }
      | c = $y - z = 0$
}}
{{eqn | o = >
      | r = 0
      | c = $x > 0$, $x - y = x - z > 0$
}}
{{end-eqn}}
This shows the equality conditions is {{iff}}.
$(1)$ can then be rearranged to '''Schur's inequality'''.
{{qed|lemma}}
Now, let $t$ be a positive even integer.
{{WLOG}}, we can assume that $x \ge y \ge z$.
By Pigeonhole Principle, at least $2$ of them have the same sign.
Suppose $x, y$ are positive.
Once again we consider the expression:
:$\paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z}} + z^t \paren {x - z} \paren {y - z}$
The first term is still non-negative.
The second term is non-negative, since:
:$z^t \ge 0 \quad$ Even Power is Non-Negative
:$x \ge z$, $y \ge z$
Thus we can still conclude $(1)$, which can then be rearranged to '''Schur's inequality'''.
Suppose $y, z$ are negative.
Then $-z, -y$ are positive, and $-z \le -y \le -x$.
Substituting $x, y, z$ for $-z, -y, -x$ in the above, the result follows.
{{qed}}
{{namedfor|Issai Schur|cat = Schur}}
Category:Algebra
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