Schur's Inequality

Theorem
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 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 $z$.

, 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:

Now we suppose two numbers are equal, but the other is neither the same number or $0$.

If $x = y > z > 0$:

If $x > y = z \ge 0$:

This shows the equality conditions is.

$(1)$ can then be rearranged to Schur's inequality.

Now, let $t$ be a positive even integer.

, 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.