Inequality by Van Khea

The theorem of inequality

April 30, 2010

Let $a_1, a_2, ...a_{m-1}, a_m, a_{m+1},..., a_{n-1}, a_n$ be positive numbers and $m\leq n-1 ; m,n\in N^{*}$ so can state as:

$\displaystyle For: m = 1\Longrightarrow \frac{a_1}{a_2}+\frac{a_2}{a_3}+...+\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}\geq n$

$\displaystyle For: m = 2\Longrightarrow\frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+...+\frac{a_{n-1}}{a_n+a_1}+\frac{a_n}{a_1+a_2}\geq \frac{n}{2}$

$\displaystyle For: m = 3\Longrightarrow\frac{a_1}{a_2+a_3+a_4}+\frac{a_2}{a_3+a_4+a_5}+...+\frac{a_{n-1}}{a_n+a_1+a_2}+\frac{a_n}{a_1+a_2+a_3}\geq \frac{n}{3}$

………………………………………………………………………………………………………….

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$For: m = m\leq n-1\Longrightarrow$ គេបាន

$\displaystyle \frac{a_1}{a_2+a_3+...+a_{m+1}}+\frac{a_2}{a_3+a_4+...+a_{m+2}}+...+\frac{a_{n-1}}{a_n+a_1+...+a_{m-1}}+\frac{a_n}{a_1+a_2+...+a_m}$

$\displaystyle \geq \frac{n}{m}$

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Inequality 08

April 29, 2010

Let function $\displaystyle f_n(x)=\biggl(1+\frac{1}{n^2}.f_{n-1}(x)\biggl)^n ; n\geq 1$ and $f_0(x)=e^x$

Prove that for $a\geq x\geq 0$ we have : $\displaystyle f(n!a)<e^a+2+\frac{1}{n!}$

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Inequality 07

April 29, 2010

Let a, b, x, y be positive real numbers satisfy two conditions below:

$1) : 1\leq x\leq a$ and $1\leq y\leq b$

$\displaystyle 2): \frac{x}{a}\leq \frac{y}{b}$

Prove that: $(x+y)(a^x+b^y)\leq (a+b)(x^a+y^b)$

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Inequality 06

April 29, 2010

Prove that for $n\geq 0$

$\displaystyle \Longrightarrow e^{\biggl(1+\frac{1}{n}\biggl)^n}\geq \biggl(1+\frac{1}{n}\biggl)^{ne}$

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Inequality 05

April 29, 2010

Let a, b, c be positive real numbers for which $a\neq b\neq c$ . Prove that :

$\displaystyle \frac{a}{b}.\frac{c-a}{c-b}+\frac{b}{c}.\frac{a-b}{a-c}+\frac{c}{a}.\frac{b-c}{b-a}+\frac{b}{a}.\frac{c-b}{c-a}+\frac{c}{b}.\frac{a-c}{a-b}+\frac{a}{c}.\frac{b-a}{b-c}\leq 3$

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Inequality 04

April 29, 2010

Let a , b , c be positive real numbers for which $a\leq b\leq c$ and $a^2+b^2+c^2=2$ . Prove that:

$\displaystyle \frac{1-a^2}{c(a+b)}+\frac{1-b^2}{a(b+c)}+\frac{1-c^2}{b(c+a)}\geq \frac{3}{4}$

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The theorem

April 29, 2010

If functions f and g are both continuous on the closed interval [a,c], and differentiable on the open interval (a, c), then there exists some x ∈ (a,b), and y∈ (b,c) ; $a\leq b\leq c$ such that

$\displaystyle \frac{(c-b)f(a)-(c-a)f(b)+(b-a)f(c)}{(c-b)g(a)-(c-a)g(b)+(b-a)g(c)}=\frac{f'(y)-f'(x)}{g'(y)-g'(x)}$

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Inequality 03

April 29, 2010

For a, b, c are some non – negative that $a\leq b\leq c$ . Prove that:

$\displaystyle \frac{a^6}{b^6}+\frac{b^6}{c^6}+\frac{c^6}{a^6}\geq \frac{1}{9}\biggl(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\biggl)\biggl(\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}\biggl)\biggl(\frac{b^3}{a^3}+\frac{c^3}{b^3}+\frac{a^3}{c^3}\biggl)$

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Inequality 02

April 29, 2010

For a, b, c are some non – negative that $a\leq b\leq c$ . Prove that :

$\displaystyle \frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\geq \frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}$

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Inequality 01

April 29, 2010

For a , b, c are some non-neqative $a\leq b\leq c$ . Prove that:

$\displaystyle 3.\biggl(\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\biggl)^2\geq \biggl(\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}\biggl)^2$ .

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The new inequality by Van Khea

April 5, 2010

For a real convex fruction $f:\mathbb{R}\longrightarrow \mathbb{R}_0^{+}$ numbers $a, b, c$ in its domain $a\leq b\leq c$ , and positive $m\geq n, p\geq n$ so we can write that:

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

For a real concave fruction $f:\mathbb{R}\longrightarrow \mathbb{R}_0^{+}$ numbers $a, b, c$ in its domain $a\leq b\leq c$ , and positive $m\leq n, p\leq n$ so we can write that:

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

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Hello world!

April 5, 2010

Welcome to WordPress.com. This is your first post. Edit or delete it and start blogging!

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Inequality by Van Khea

The theorem of inequality

Inequality 08

Inequality 07

Inequality 06

Inequality 05

Inequality 04

The theorem

Inequality 03

Inequality 02

Inequality 01

The new inequality by Van Khea

For a real convex fruction $f:\mathbb{R}\longrightarrow \mathbb{R}_0^{+}$ numbers $a, b, c$ in its domain $a\leq b\leq c$ , and positive $m\geq n, p\geq n$ so we can write that:

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

For a real concave fruction $f:\mathbb{R}\longrightarrow \mathbb{R}_0^{+}$ numbers $a, b, c$ in its domain $a\leq b\leq c$ , and positive $m\leq n, p\leq n$ so we can write that:

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

Hello world!

van khea

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For a real convex fruction numbers in its domain , and positive so we can write that:

For a real concave fruction numbers in its domain , and positive so we can write that:

van khea

Calendar

Categories

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Meta

For a real convex fruction $f:\mathbb{R}\longrightarrow \mathbb{R}_0^{+}$ numbers $a, b, c$ in its domain $a\leq b\leq c$ , and positive $m\geq n, p\geq n$ so we can write that:

For a real concave fruction $f:\mathbb{R}\longrightarrow \mathbb{R}_0^{+}$ numbers $a, b, c$ in its domain $a\leq b\leq c$ , and positive $m\leq n, p\leq n$ so we can write that: