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

(Van Khea):Problem 026

Filed under: Uncategorized — KKKVVV @ 9:08 am

Let a, b, c, k>0; a\leq b\leq c. Prove that:

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

Proof

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

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

From the conditional a\leq b\leq c ; k>0\Longrightarrow a^k\leq b^k\leq c^k

From the inequality by Van Khea 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

\displaystyle \Longrightarrow \frac{c^k-b^k}{a}-\frac{c^k-a^k}{b}+\frac{b^k-a^k}{c}\geq 0

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

Therefor the proof is completed.

See also the inequality by VanKhea

(Van Khea):Problem 025

Filed under: Uncategorized — KKKVVV @ 8:55 am

Let a_1, a_2, ..., a_n be positive real numbers satisfying a_1^{n-1}+a_2^{n-1}+...+a_n^{n-1}=1. Prove that:

a_1+a_2+...+a_n\geq n^2a_1a_2...a_n