July | 2010 | 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

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

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

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.

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:

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

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July 29, 2010

(van khea):Problem 023

Filed under: Uncategorized — KKKVVV @ 1:19 am

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

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

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(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:

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

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(van khea):Problem 021

Filed under: Uncategorized — KKKVVV @ 1:03 am

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

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

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(van khea):Problem 020

Filed under: Uncategorized — KKKVVV @ 12:57 am

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

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

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(van khea):Problem 019

Filed under: Uncategorized — KKKVVV @ 12:53 am

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

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

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(Van Khea):Problem 018

Filed under: Uncategorized — KKKVVV @ 12:45 am

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

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

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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:

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

Proof

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

$f''(x)=k(k-1)x^{k-2}>0$

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.

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:

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

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(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:

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

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(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:

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

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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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July 31, 2010

Let be positive real numbers and satisfying . Prove that for we have:

Van khea

Proof

We have

Let

From the inequality by VanKhea for and we get

Let . Prove that:

Proof

Let

From the conditional

From the inequality by Van Khea for ; and we get:

Therefor the proof is completed.

Let be positive real numbers satisfying . Prove that:

July 30, 2010

Let . Prove that for we have:

July 29, 2010

Let . Prove that:

Let . Prove that:

Let . Prove that:

Let . Prove that:

Let . Prove that:

Let . Prove that:

July 28, 2010

Let . Prove that:

Proof

Let so we need to prove that

From Cauchy’s inequality we have

Therefor with the inequality is trued.

Now we suppose that

So the inequality above became

From the conditional we had so we can rewrite

We have

Let

Let

From the inequality by van khea for we get

Therefor the proof is completed.

July 27, 2010

Let . Prove that:

Let . Prove that:

Let . Prove that:

July 26, 2010

Let . Prove that for we have:

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

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

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

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

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

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

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:

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:

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

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

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:

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

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

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

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

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

Let $k=1$ so we need to prove that

Therefor with $k=1$ the inequality is trued.

Now we suppose that $k>1$

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

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

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

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

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

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:

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