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.

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April 2, 2011

Van Khea’s inequality and application

Filed under: Van Khea 01 — KKKVVV @ 9:37 pm

Theorem:

1) For a real convex function f(x), numbers a, b, c in its domain such that a\leq b\leq c and let m, n, p are positive real numbers such that m\geq n ; p\geq n then we have

m(c-b)f(a)-n(c-a)f(b)+p(b-a)f(c)\geq 0

Special case if m=n=p\neq 0 then we have (c-b)f(a)-(c-a)f(b)+(b-a)f(c)\geq 0

2) For a real concave function f(x), numbers a, b, c in its domain such that a\leq b\leq c and let m, n, p are positive real numbers such that m\leq n ; p\leq n then we have

m(c-b)f(a)-n(c-a)f(b)+p(b-a)f(c)\leq 0

Special case if m=n=p\neq 0 then we have (c-b)f(a)-(c-a)f(b)+(b-a)f(c)\leq 0

Proof

Suppose that c_1\in (a, b) ; c_2\in (b, c) then from Lagrange’s theorem we get:

\displaystyle f'(c_1)=\frac{f(b)-f(a)}{b-a} ; f'(c_2)=\frac{f(c)-f(b)}{c-b}

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