April 16, 2011

A generalization of Schur’s inequality

Filed under: a generalization of Schur's inequality — KKKVVV @ 12:29 am

(Theorem): Consider a, b, c, x, y, z\in R where a\geq b\geq c and either x\ge y\geq z or x\leq y\leq z. Let k\in Z^{+} and let f:\mathbb{R}\rightarrow \mathbb{R}_{0}^{+} be either convex or monotonic, then

(a-b)^k(a-c)^kf(x)+(b-a)^k(b-c)^kf(y)+(c-a)^k(c-b)^kf(z)\geq 0


There are two cases:

+ If k=2n' ; n'\in N then the above problem is alway true.

+ If k=2n'+1 ; n'\in N then we can write as:


Now we need to prove that: (a-b)^k(a-c)^kf(x)-(a-b)^k(b-c)^kf(y)+(a-c)^k(b-c)^kf(z)\geq 0

Without loss of generality, suppose that x\leq y\leq z then:

Let \displaystyle m=\frac{(a-b)^k(a-c)^k}{z-y} ; n=\frac{(a-b)^k(b-c)^k}{z-x} ; p=\frac{(a-c)^k(b-c)^k}{y-x}

then we have m\geq n\geq 0; p\geq n\geq 0

From van khea’s inequality with convex funtion f we have

m(z-y)f(x)-n(z-x)f(y)+p(y-x)f(z)\geq 0

\displaystyle \Leftrightarrow \frac{(a-b)^k(a-c)^k}{z-y}(z-y)f(x)-\frac{(a-b)^k(b-c)^k}{z-x}(z-x)f(y)+\frac{(a-c)^k(b-c)^k}{y-x}(y-x)f(z)\geq 0

\Rightarrow(a-b)^k(a-c)^kf(x)-(a-b)^k(b-c)^kf(y)+(a-c)^k(b-c)^kf(z)\geq 0

Therefore the proof is completed.