INEQUALITY'S BLOG

new inequality

Theoreom:

(Van Khea):Let a, b, x, y>0, \alpha\geq 1, \beta, \gamma \geq 0 ; \alpha-\beta-\gamma=1. Prove that:

\displaystyle \frac{a^{\alpha}}{x^{\beta}p^{\gamma}}+\frac{b^{\alpha}}{y^{\beta}q^{\gamma}}\geq \frac{(a+b)^{\alpha}}{(x+y)^{\beta}(p+q)^{\gamma}}

Let x_i, y_i, z_i>0 ; i=1, 2, ..., n and \alpha\geq 1, \beta, \gamma \geq 0 such that \alpha -\beta -\gamma=1. We get:

\displaystyle \frac{x_1^{\alpha}}{y_1^{\beta}z_1^{\gamma}}+\frac{x_2^{\alpha}}{y_2^{\beta}z_2^{\gamma}}+...+\frac{x_n^{\alpha}}{y_n^{\beta}z_n^{\gamma}}\geq \frac{(x_1+x_2+...+x_n)^{\alpha}}{(y_1+y_2+...+y_n)^{\beta}(z_1+z_2+...+z_n)^{\gamma}}

Let \{x_{i,j}\}>0 ; (i=1, 2, ..., m ; j=1, 2, ...,=n) and for a_1\geq 1; a_2, a_3,...,a_n\geq 0 such that: a_1-a_2-...-a_n=1. So we get:

\displaystyle \frac{x_{1,1}^{a_1}}{\prod_{j=2}^{n}x_{1,j}^{a_j}}+...+\frac{x_{m,1}^{a_1}}{\prod_{j=2}^{n}x_{m,j}^{a_j}}\geq \frac{(x_{1,1}+...+x_{m,1})^{a_1}}{\prod_{j=2}^{n}(x_{1,j}+...+x_{m,j})^{a_j}}

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:

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

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)



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