# INEQUALITY'S BLOG

## 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

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

Proof

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:

$(a-b)^k(a-c)^kf(x)+(b-a)^k(b-c)^kf(y)+(c-a)^k(c-b)^kf(z)$$=(a-b)^k(a-c)^kf(x)-(a-b)^k(b-c)^kf(y)+(a-c)^k(b-c)^kf(z)$

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.