In mathematics, Schur's inequality, named after Issai Schur, establishes that for all non-negative real numbers x, y, z, and t>0,
![{\displaystyle x^{t}(x-y)(x-z)+y^{t}(y-z)(y-x)+z^{t}(z-x)(z-y)\geq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb05031b3c81fe951be7c1ab07bf093b357ccc62)
with equality if and only if x = y = z or two of them are equal and the other is zero. When t is an even positive integer, the inequality holds for all real numbers x, y and z.
When
, the following well-known special case can be derived:
![{\displaystyle x^{3}+y^{3}+z^{3}+3xyz\geq xy(x+y)+xz(x+z)+yz(y+z)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/559084c7cc0e9215d9c4dbb3e6088a280d340031)
Proof
Since the inequality is symmetric in
we may assume without loss of generality that
. Then the inequality
![{\displaystyle (x-y)[x^{t}(x-z)-y^{t}(y-z)]+z^{t}(x-z)(y-z)\geq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92dd049645383022315ce4a06443dd25293e3a06)
clearly holds, since every term on the left-hand side of the inequality is non-negative. This rearranges to Schur's inequality.
Extensions
A generalization of Schur's inequality is the following: Suppose a,b,c are positive real numbers. If the triples (a,b,c) and (x,y,z) are similarly sorted, then the following inequality holds:
![{\displaystyle a(x-y)(x-z)+b(y-z)(y-x)+c(z-x)(z-y)\geq 0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3829cbc30f8679aba90dbfa577bcd05c536eb1d)
In 2007, Romanian mathematician Valentin Vornicu showed that a yet further generalized form of Schur's inequality holds:
Consider
, where
, and either
or
. Let
, and let
be either convex or monotonic. Then,
![{\displaystyle {f(x)(a-b)^{k}(a-c)^{k}+f(y)(b-a)^{k}(b-c)^{k}+f(z)(c-a)^{k}(c-b)^{k}\geq 0}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5555013928e82454e1f109e5704e4ead1a9e6f42)
The standard form of Schur's is the case of this inequality where x = a, y = b, z = c, k = 1, ƒ(m) = mr.[1]
Another possible extension states that if the non-negative real numbers
with and the positive real number t are such that x + v ≥ y + z then[2]
![{\displaystyle x^{t}(x-y)(x-z)(x-v)+y^{t}(y-x)(y-z)(y-v)+z^{t}(z-x)(z-y)(z-v)+v^{t}(v-x)(v-y)(v-z)\geq 0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eebbaf730e30f78ab2baf77e2392372a20e82f08)
Notes
- ^ Vornicu, Valentin; Olimpiada de Matematica... de la provocare la experienta; GIL Publishing House; Zalau, Romania.
- ^ Finta, Béla (2015). "A Schur Type Inequality for Five Variables". Procedia Technology. 19: 799–801. doi:10.1016/j.protcy.2015.02.114.