INEQUALITY'S BLOG

July 31, 2010

Problem 027

Filed under: Uncategorized — KKKVVV @ 10:49 pm

Let a, b, c be positive real numbers and satisfying a\leq b\leq c. Prove that for k, s>0 we have:

\displaystyle \frac{a^{k}}{b^{s}}+\frac{b^{k}}{c^{s}}+\frac{c^{k}}{a^{s}}\geq \frac{b^{k}}{a^{s}}+\frac{c^{k}}{b^{s}}+\frac{a^{k}}{c^{s}}

 

Van khea   

Proof

We have a\leq b\leq c\Longrightarrow a^k\leq b^k\leq c^k ; k>0

Let \displaystyle f(x)=\frac{1}{x^{\frac{s}{k}}}; k, s>0

\displaystyle f''(x)=\frac{s}{k}(\frac{s}{k}+1)x^{-(\frac{s}{k}+2)}>0

 From the inequality by VanKhea for m=n=p\neq 0; a^k\leq b^k\leq c^k and f''(x)>0 we get

(c^k-b^k)f(a^k)-(c^k-a^k)f(b^k)+(b^k-a^k)f(c^k)\geq 0

\Longrightarrow\displaystyle \frac{a^{k}}{b^{s}}+\frac{b^{k}}{c^{s}}+\frac{c^{k}}{a^{s}}\geq \frac{b^{k}}{a^{s}}+\frac{c^{k}}{b^{s}}+\frac{a^{k}}{c^{s}}

See also the inequality by Van Khea

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