# INEQUALITY'S BLOG

## October 20, 2010

### Van Khea’s inequality

Filed under: the new math — KKKVVV @ 2:48 pm

## Theoreom:

Let $a_1, a_3, a_3, b_1, b_2, b_3, c_1, c_2, c_3, x_1, x_2, x_3$ be positive real numbers and let

$y_1, y_2, y_3\in (0; 1)$. For all $p, q, r, s, t$ be real numbers satisfying $p\geq 1; q\geq 0; r\geq 0; s\geq 0; t\geq 0$ and $p-q-r-s-t=1$. So we have:

$\displaystyle \frac{x_1^{p}(1+y_1)^{q}}{a_1^{r}b_1^{s}c_1^{t}}+\frac{x_2^{p}(1+y_2)^{q}}{a_2^{r}b_2^{s}c_2^{t}}+\frac{x_3^{p}(1+y_3)^{q}}{a_3^{r}b_3^{s}c_3^{t}}$

$\displaystyle \geq \biggl(\frac{1+\sqrt[3]{y_1y_2y_3}}{3}\biggl)^{q}.\frac{(x_1+x_2+x_3)^{p}}{(a_1+a_2+a_3)^{r}(b_1+b_2+b_3)^{s}(c_1+c_2+c_3)^{t}}$

Equality hold when: $a_1=a_2=a_3, b_1=b_2=b_3, c_1=c_2=c_3; x_1=x_2=x_3; y_1=y_2=y_3$