INEQUALITY'S BLOG

October 17, 2012

Last two digit (vankhea formula)

Filed under: the new math — KKKVVV @ 11:25 am

If $p\equiv r(mod 20); r\in \mathbb{N}^{*}; 1\leq r\leq 20$ and $\forall{x, y\in \mathbb{N}}; 0\leq x\leq 9; 0\leq y\leq 9$. Then we get: $(\overline{xy})^p\equiv y^p+10x(y^r)'(mod 100)$ with $(y^r)'=ry^{r-1}$

Last one digit (vankhea formula )

Filed under: the new math — KKKVVV @ 11:18 am

If $p\equiv r(mod 4) ; r\in \mathbb{N}^{*} ; 1\leq r\leq 4$ and $\forall{x\in \mathbb{N}} ; 0\leq x\leq 9$ we get: $x^p\equiv x^r(mod 10)$

December 15, 2010

THE INEQUALITY 03: VAN KHEA

Filed under: the new math,Van Khea 03 — KKKVVV @ 2:41 pm

A GENERALIZATION OF VANKHEA’S INEQUALITY

Filed under: the new math,Van Khea 04 — KKKVVV @ 10:03 am

December 13, 2010

Inequality

Filed under: the new math — KKKVVV @ 3:12 pm

November 30, 2010

van khea

Filed under: the new math — KKKVVV @ 10:26 pm

Let $a, b, c, r, s$ be positive real numbers such that $a\leq b\leq c$ or $a\geq b\geq c$. Prove that: $a^{r+s}+b^{r+s}+c^{r+s}\geq a^rb^s+b^rc^s+c^ra^s$

November 26, 2010

Problem 72: van khea

Filed under: the new math — KKKVVV @ 8:52 am

Let $a, b, c$ be positive real numbers such that $a\geq b\geq c$ and $abc=1$. Prove that: $\displaystyle \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$

Problem 71: van khea

Filed under: the new math — KKKVVV @ 8:50 am

Let $a, b, c$ be positive real numbers such that $a\leq b\leq c$. Prove that: $\displaystyle \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq \sqrt{\frac{3}{2}(a+b+c)}$

November 22, 2010

Problem 70: van khea

Filed under: the new math — KKKVVV @ 2:40 pm

Prove that for all $a, b, c\geq 0$ we have: $a^4+b^4+c^4+(a^2+b^2+c^2)^2\geq 4(a^3b+b^3c+c^3a)$

November 5, 2010

Problem 58 (Van Khea)

Filed under: the new math — KKKVVV @ 10:37 pm

Let $a, b, c>0 ; ab+bc+ca=1$. Prove that: $\displaystyle a^{\frac{5}{2}}\sqrt{b+c}+b^{\frac{5}{2}}\sqrt{c+a}+c^{\frac{5}{2}}\sqrt{a+b}\leq \frac{\sqrt{2}}{\sqrt{3}}(a^3+b^3+c^3)^{\frac{2}{3}}$

Problem 57 (Van Khea)

Filed under: the new math — KKKVVV @ 7:32 pm

Let $a, b, c>0 ; a+b+c=3$. Prove that: $\displaystyle \frac{a^{\frac{2}{3}}}{b\sqrt{b+2c}}+\frac{b^{\frac{2}{3}}}{c\sqrt{c+2a}}+\frac{c^{\frac{2}{3}}}{a\sqrt{a+2b}}\geq \sqrt{3}$

Inequality: Problem 56

Filed under: the new math — KKKVVV @ 5:05 pm

(Van Khea): Let $a\geq b\geq c>0 ; abc=1$. Prove that: $\displaystyle \frac{a}{b}\sqrt{b+c}+\frac{b}{c}\sqrt{c+a}+\frac{c}{a}\sqrt{a+b}\leq \frac{\sqrt{2}}{\sqrt{3}}\biggl(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\biggl)^2$

Inequality: Problem 55

Filed under: the new math — KKKVVV @ 4:57 pm

(Van Khea): Let $a, b, c>0 ; abc=1$. Prove that: $\displaystyle (ab^2+bc^2+ca^2)\sqrt{a^2b+b^2c+c^2a}\geq \frac{\sqrt{3}}{\sqrt{2}}(ab^2\sqrt{c+a}+bc^2\sqrt{a+b}+ca^2\sqrt{b+c})$

Inequality: Problem 54

Filed under: the new math — KKKVVV @ 1:53 pm

(Van Khea): Let $a, b, c>0 ; a+b+c=a^2+b^2+c^2$.

Prove that: $\displaystyle \frac{a^{\frac{2}{3}}}{\sqrt{b+c}}+\frac{b^{\frac{2}{3}}}{\sqrt{c+a}}+\frac{c^{\frac{2}{3}}}{\sqrt{a+b}}\geq \frac{\sqrt{3}}{\sqrt{2}}\sqrt{a+b+c}$

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