# INEQUALITY'S BLOG

## Theoreom:

(Van Khea) :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: $a_1=a_2=a_3, b_1=b_2=b_3, c_1=c_2=c_3; x_1=x_2=x_3; y_1=y_2=y_3$

## Theoreom:

(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:

## Theoreom:

(Van Khea): Let $a, b, c$ be positive real numbers and satisfying $a\leq b\leq c$ . So we have: $b^{c-a}\geq a^{c-b}.c^{b-a}$

## Theoreom:

(Van Khea): Let $a, b, c\in [-1, 1]$. So we have

$(1-abc)^3\geq (1+abc)(1-a^2)(1-b^2)(1-c^2)$

## Theoreom:

(Van Khea): Let $x_i, y_i \forall{i=1, 2, ... , n}$  be positive real numbers and for all $a_i \geq 0 \forall{i=1, 2, ... , n}$ satisfy $a_1+a_2+...+a_n=1$ . So we have:

$\displaystyle \sum_{i=1}^{n}a_ix_iy_i\leq \biggl(\sum_{i=1}^{n}a_ix_i^p\biggl)^{\frac{1}{p}}\biggl(\sum_{i=1}^{n}a_iy_i^q\biggl)^{\frac{1}{q}}$ $\displaystyle \forall{p, q>0}$ និង $\displaystyle \frac{1}{p}+\frac{1}{q}\leq 1$

## Theorem:

(Van Khea) :1/ Let a real convex function $f(x)$ for all $x_1, x_2, ..., x_n$ in its domain. Suppose that for all positive real numbers $a_1, a_2, ..., a_n$ satisfies $a_1x_1-a_2x_2-...-a_nx_n$ in its domain and $a_1-a_2-...-a_n=1$. Then we have:

$f(a_1x_1-a_2x_2-...-a_nx_n)\geq a_1f(x_1)-a_2f(x_2)...-a_nf(x_n)$.

2/ Let a real concave function $f(x)$ for all $x_1, x_2, ..., x_n$ in its domain. Suppose that for all positive real numbers $a_1, a_2, ..., a_n$ satisfies $a_1x_1-a_2x_2-...-a_nx_n$ in its domain and $a_1-a_2-...-a_n=1$. Then we have:

$f(a_1x_1-a_2x_2-...-a_nx_n)\leq a_1f(x_1)-a_2f(x_2)...-a_nf(x_n)$.

## Theorem:

### (Van Khea): Let $f:R\longrightarrow R^{+}$ is a convex function. For any positive real numbers $m, n, p$ that $m\geq n , p\geq n$ and $m-n+p=1$. So for $a\leq b\leq c$ we can stat as:

$mf(a)-nf(b)+pf(c)\geq f(ma-nb+pc)$

Special case:

1. If $m=n=p=1$ $\Longrightarrow f(a)-f(b)+f(c)\geq f(a-b+c)$
2. If $n=0$ then $mf(a)+pf(c)\geq f(ma+pc)$

## Theorem:

(van khea): 1/ Let $a, b, c$ be positive real numbers such that $a\leq b\leq c$ . So we have:

$\displaystyle \frac{a^t+b^t-c^t}{a^s}+\frac{b^t+c^t-a^t}{b^s}+\frac{c^t+a^t-b^t}{c^s}\leq a^{t-s}+b^{t-s}+c^{t-s} ; \forall{s, t\geq 0}$

(van khea):2/  Let $a, b, c$ be positive real numbers such that $a\geq b\geq c$ . So we have:

$\displaystyle \frac{a^t+b^t-c^t}{a^s}+\frac{b^t+c^t-a^t}{b^s}+\frac{c^t+a^t-b^t}{c^s}\geq a^{t-s}+b^{t-s}+c^{t-s} ; \forall{s, t\geq 0}$

## Theoreom:

(van khea): Let $a, b, c$ be positive real numbers such that:  $a\leq b\leq c$ or $a\geq b\geq c$ .Then

$\displaystyle \frac{a^t+b^t-c^t}{ab}+\frac{b^t+c^t-a^t}{bc}+\frac{c^t+a^t-b^t}{ca}$$\leq a^{t-2}+b^{t-2}+c^{t-2}$$;\forall{t\geq 1}$

## Theoreom:

(van khea): Let $a, b, c$ be positive real numbers such that:  $a\leq b\leq c$ or $a\geq b\geq c$ .Then

$(a^r-b^r)(a^s-c^s)a^t+(b^r-a^r)(b^s-c^s)b^t+(c^r-a^r)(c^s-b^s)c^t$$\geq 0 ; \forall{r, s, t\geq 0}$

## Theoreom:

(van khea): Consider $a, b, c\in R ; x, y, z\in I$ where $a\geq b\geq c$ and $x\leq y\leq z$ or $x\geq y\geq z$. Let $f:\mathbb{R}\longrightarrow \mathbb{R}_{0}^{+}$ be convex function or momotonic and let $k, l \geq 0$ where $k\&l$ in the same neture. Then:

$(a^k-b^k)(a^l-c^l)f(x)+(b^k-a^k)(b^l-c^l)f(y)+(c^k-a^k)(c^l-b^l)f(z)$$\geq 0$

## Theoreom:

(van khea): Consider $a, b, c\in R ; x, y, z\in I$ where $a\geq b\geq c$ and $x\leq y\leq z$ or $x\geq y\geq z$. Let $f:\mathbb{R}\longrightarrow \mathbb{R}_{0}^{+}$ be convex function or momotonic and let $k, l\in \mathbb{Z^{+}}$ where $k\&l$ in the same neture. Then:
$(a-b)^k(a-c)^lf(x)+(b-a)^k(b-c)^lf(y)+(c-a)^k(c-b)^lf(z)\geq 0$

## Theorem:

(van khea): Let $a, b, c, r, s, t\geq 0$ then we have:

$\displaystyle \sum_{cyc}a^{r+s+t}+\sum_{cyc}a^rb^sc^t\geq \sum_{cyc}a^rb^s(a^t+b^t)$