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   


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}


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 0a^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

July 30, 2010

(Van Khea): Problem 024

Filed under: Uncategorized — KKKVVV @ 3:09 am

Let a, b, c>0; a^2+b^2+c^2=1. Prove that for k\geq 0 we have:

\displaystyle a(bc)^{\frac{1}{4.3^{k}}}+b(ca)^{\frac{1}{4.3^{k}}}+c(ab)^{\frac{1}{4.3^{k}}}\leq 3^{\frac{2.3^{k}-1}{4.3^{k}}}

July 29, 2010

(van khea):Problem 023

Filed under: Uncategorized — KKKVVV @ 1:19 am

Let x, y, z>0; xyz=1. Prove that:

\displaystyle \frac{\biggl(\frac{1}{x^{2}}+\frac{1}{y}\biggl)^{2011}}{x(y+z)}+\frac{\biggl(\frac{1}{y^{2}}+\frac{1}{z}\biggl)^{2011}}{y(z+x)}+\frac{\biggl(\frac{1}{z^{2}}+\frac{1}{x}\biggl)^{2011}}{z(x+y)}\geq 3.2^{2010}

(van khea):Problem 022

Filed under: Uncategorized — KKKVVV @ 1:11 am

Let \displaystyle x, y, z>0; 2(xy+yz+zx)\geq \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}. Prove that:

\displaystyle \frac{(x+y)^{3}}{1+2xy}+\frac{(y+z)^{3}}{1+2yz}+\frac{(z+x)^{3}}{1+2zx}\geq 8

(van khea):Problem 021

Filed under: Uncategorized — KKKVVV @ 1:03 am

Let a, b, c>k\geq 2. Prove that:

\displaystyle \frac{1}{\sqrt[3]{a-k}}+\frac{1}{\sqrt[3]{b-k}}+\frac{1}{\sqrt[3]{c-k}}+\sqrt[3]{abc}\geq k+4

(van khea):Problem 020

Filed under: Uncategorized — KKKVVV @ 12:57 am

Let a, b, c>0 ; a+b+c=1. Prove that:

5(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq \sqrt{ab+bc+ca}+24\sqrt[3]{abc}

(van khea):Problem 019

Filed under: Uncategorized — KKKVVV @ 12:53 am

Let a, b, c>0. Prove that:

a\sqrt{a+2b}+b\sqrt{b+2c}+c\sqrt{c+2a}\displaystyle \geq \frac{11(ab+bc+ca)+48\sqrt[3]{a^{2}b^{2}c^{2}}}{9\sqrt{a+b+c}}

(Van Khea):Problem 018

Filed under: Uncategorized — KKKVVV @ 12:45 am

Let x, y, z>0 ; xy+yz+zx=xyz. Prove that:

\displaystyle \frac{xy}{(1+x)(1+y)}+\frac{yz}{(1+y)(1+z)}+\frac{zx}{(1+z)(1+x)}\geq \frac{27}{16}

July 28, 2010

(Van Khea):Problem 017

Filed under: the new math — KKKVVV @ 5:26 pm

Let a\geq b\geq c>0. Prove that:

\displaystyle \frac{a^{k}b}{c}+\frac{b^{k}c}{a}+\frac{c^{k}a}{b}\geq a^k+b^k+c^k ; \forall{k\geq 1}


Let k=1 so we need to prove that

\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+b+c

From Cauchy’s inequality we have

\displaystyle \frac{ab}{c}+\frac{bc}{a}\geq 2\sqrt{\frac{ab^2c}{ca}}=2b

\displaystyle \frac{bc}{a}+\frac{ca}{b}\geq 2c

\displaystyle \frac{ca}{b}+\frac{ab}{c}\geq 2a

\displaystyle \Longrightarrow 2(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b})\geq 2(a+b+c)

\displaystyle \Longrightarrow \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+b+c

Therefor with k=1 the inequality is trued.

Now we suppose that k>1

So the inequality above became

\displaystyle \frac{a^kb}{c}-a^k+\frac{b^kc}{a}-b^k+\frac{c^ka}{b}-c^k\geq 0

From the conditional we had a\geq b\geq c>0 so we can rewrite

\displaystyle ca(a-b)c^k-bc(a-c)b^k+ab(b-c)a^k\geq 0

We have a\geq b\geq c\Longrightarrow ab\geq ca\geq bc

Let m=ca ; n=bc; p=ab\Longrightarrow m\geq n; p\geq n

Let f(x)=x^k ; x>0 ; k>1


From the inequality by van khea for a\geq b\geq c>0; m\geq n; p\geq n; f''(x)>0 we get

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

\displaystyle \Longrightarrow ca(a-b)c^k-bc(a-c)b^k+ab(b-c)a^k\geq 0

Therefor the proof is completed.

See also the inequality by Van Khea 😀

July 27, 2010

(Van Khea):Problem 016

Filed under: the new math — KKKVVV @ 5:19 pm

Let a, b, c>0 ;a^2+b^2+c^2=1. Prove that:

\displaystyle \frac{a^{3}}{bc}+\frac{b^{3}}{ca}+\frac{c^{3}}{ab}+k.\frac{a^{5}+b^{5}+c^{5}}{abc(a+b+c)}\geq k+3 ; \forall{k\geq 3}

(Van Khea):Problem 015

Filed under: the new math — KKKVVV @ 5:11 pm

Let a, b, c>0 ; a^2+b^2+c^2=1. Prove that:

\displaystyle \frac{bc}{a^{4}}+\frac{ca}{b^{4}}+\frac{ab}{c^{4}}+\biggl(\frac{1}{a}+\frac{1}{b}\biggl)\biggl(\frac{1}{b}+\frac{1}{c}\biggl)\biggl(\frac{1}{c}+\frac{1}{a}\biggl)\geq 3(8\sqrt{3}+3)

(van Khea):Problem 014

Filed under: the new math — KKKVVV @ 4:59 pm

Let a, b, c>0 ; a^2+b^2+c^2=1. Prove that:

\displaystyle ab+bc+ca+\frac{k}{abc}\geq 3k\sqrt{3}+1\displaystyle \forall{k\geq \frac{1}{3\sqrt{3}}}

July 26, 2010

(Van Khea):Problem 013

Filed under: the new math — KKKVVV @ 8:47 pm

Let a, b, c, x, y, z>0; x^2+y^2+z^2=1. Prove that for \displaystyle k\geq \frac{a+b+c}{9} we have:

\displaystyle ax+by+cz+\frac{k}{xyz}\geq (3k+\frac{a+b+c}{3})\sqrt{3}

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