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}

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!}

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)

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}

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

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}

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)}

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)

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}

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.

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

Hello world!

April 5, 2010

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