# 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:
$x^p\equiv x^r(mod 10)$

## August 29, 2012

### vankhea 2010.12 inequality

Filed under: Van Khea 06 — KKKVVV @ 10:43 am

Let $a, b, c, p, q$ be positive real numbers such that $abc=1$. Prove that
$\displaystyle \frac{a^m}{(pb+qc)^n}+\frac{b^m}{(pc+qa)^n}+\frac{c^m}{(pa+qb)^n}\geq \frac{3}{(p+q)^n}$
with $m>0 ; n\in R$;$m\geq n$ satisfy $\displaystyle mln3+nln \frac{4}{3}\geq ln \frac{9}{4}$

## May 18, 2012

### Problem 315 vankhea

Filed under: Problem by Van Khea — KKKVVV @ 11:28 am

if $a, b, c$ are positive real numbers satisfying $ab+bc+ca=3$ then prove that
$(2+3a^3)(2+3b^3)(2+3c^3)\geq 125$

## May 17, 2012

### Problem 313 vankhea

Filed under: Problem by Van Khea — KKKVVV @ 11:03 pm

If $a, b, c, d$ are positive real numbers then
$(3+a^3)(3+b^3)(3+c^3)(3+d^3)\geq 4(a+b+c+d)^3$

## December 8, 2011

### Problem 309 van khea

Filed under: Problem by Van Khea — KKKVVV @ 6:48 am

If $a, b, c$ are positive real numbers such that $a+b+c=3$. Prove that
$\sqrt{1+a^4}+\sqrt{1+b^4}+\sqrt{1+c^4}\leq \sqrt{2}(a^2+b^2+c^2)$

## December 5, 2011

### Problem 307 vankhea

Filed under: Problem by Van Khea — KKKVVV @ 7:13 pm

if $a, b, c$ are positive real numbers. Prove that:
$\displaystyle (a^3+b^3)^2\geq \frac{1}{2}(a^4+b^4)(a+b)^2$

## October 3, 2011

### Problem 307 Van Khea

Filed under: Problem by Van Khea — KKKVVV @ 7:13 pm

if $a, b, c$ are positive real numbers such that $abc=1$. Prove that:

$\displaystyle \frac{1}{a\sqrt{1+3b}}+\frac{1}{b\sqrt{1+3c}}+\frac{1}{c\sqrt{1+3a}}\geq \frac{3}{2}$

## September 30, 2011

### Problem 306 Van Khea

Filed under: Problem by Van Khea — KKKVVV @ 2:00 am

If $a, b, c$ are positive real numbers such that $abc=1$. Prove that:

$\displaystyle \sqrt{\frac{a}{b(c+3)}}+\sqrt{\frac{b}{c(a+3)}}+\sqrt{\frac{c}{a(b+3)}}\geq \frac{3}{2}$

## September 1, 2011

### Problem 305 Van Khea

Filed under: Problem by Van Khea — KKKVVV @ 2:33 am

If $a, b, c$ are positive real numbers such that $a+b+c=3$. Prove that

$29(a^2+b^2+c^2)+30abc\geq 6(ab^2+bc^2+ca^2)+99$

Proof
Using the well-known identity $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=9-2(ab+bc+ca)$
Then the inequality above equivalent to
$29(ab+bc+ca)+3(ab^2+bc^2+ca^2)\leq 81+30abc$
$\Leftrightarrow 29.3(ab+bc+ca)+9(ab^2+bc^2+ca^2)\leq 243+90abc$
$\Leftrightarrow 29(a+b+c)(ab+bc+ca)+9(ab^2+bc^2+ca^2)\leq 243+90abc$
$\Leftrightarrow 29(a^2b+b^2c+c^2a)+38(ab^2+bc^2+ca^2)+42abc\leq 243$
Let $\displaystyle a=\frac{3x}{x+y+z}; b=\frac{3y}{x+y+z}; c=\frac{3z}{x+y+z}; \forall{x, y, z>0}$
The inequality transforms into
$29(x^2y+y^2z+z^2x)+38(xy^2+yz^2+zx^2)+42xyz\leq 9(x+y+z)^3$
$\Leftrightarrow 9(x^3+y^3+z^3)+12xyz\geq 3(x^2y+y^2z+z^2x)+11(xy^2+yz^2+zx^2)$
$\Leftrightarrow x(y+2z-3x)^2+y(z+2x-3y)^2+z(x+2y-3z)^2\geq 0$
Which is clear true. Equality occurs for $x=y=z\Leftrightarrow a=b=c=1$

## August 23, 2011

### Problem 304 Van Khea

Filed under: Problem by Van Khea — KKKVVV @ 12:52 pm

If $a, b, c$ are positive real numbers such that $a^2+b^2+c^2+2abc=5$ then prove that

$\displaystyle \frac{1}{a^3+b^3+c^3}+\frac{1}{abc}\geq \frac{4}{3}$

Solution
We have
$\displaystyle \frac{1}{a^3+b^3+c^3}+\frac{1}{abc}=\frac{1}{a^3+b^3+c^3}+\frac{a^2+b^2+c^2+2abc}{5abc}$
$\displaystyle \Rightarrow \frac{1}{a^3+b^3+c^3}+\frac{a}{5bc}+\frac{b}{5ca}+\frac{c}{5ab}+\frac{2}{5}\geq \frac{4}{3}$
$\displaystyle \Leftrightarrow \frac{1}{a^3+b^3+c^3}+\frac{a}{5bc}+\frac{b}{5ca}+\frac{c}{5ab}\geq \frac{14}{15}$
Let $\displaystyle a=\sqrt[3]{\frac{3x}{x+y+z}}; b=\sqrt[3]{\frac{3y}{x+y+z}}; c=\sqrt[3]{\frac{3z}{x+y+z}}$ then we get $a^3+b^3+c^3=3$
Sitting $a, b, c$ yields
$\displaystyle \frac{1}{3}+\frac{\sqrt[3]{x+y+z}}{5\sqrt[3]{3}}\biggl(\sqrt[3]{\frac{x}{yz}}+\sqrt[3]{\frac{y}{zx}}+\sqrt[3]{\frac{z}{xy}}\biggl)\geq \frac{14}{15}$
$\displaystyle \Leftrightarrow \frac{\sqrt[3]{x+y+z}}{\sqrt[3]{3}}\biggl(\sqrt[3]{\frac{x}{yz}}+\sqrt[3]{\frac{y}{zx}}+\sqrt[3]{\frac{z}{xy}}\biggl)\geq 3$
From $AM-GM$ inequality we have
$\sqrt[3]{x+y+z}\geq \sqrt[3]{3\sqrt[3]{xyz}}$ and $\displaystyle \sqrt[3]{\frac{x}{yz}}+\sqrt[3]{\frac{y}{zx}}+\sqrt[3]{\frac{z}{xy}}\geq 3\sqrt[3]{\frac{1}{\sqrt[3]{xyz}}}$
then we get
$\displaystyle \frac{\sqrt[3]{x+y+z}}{\sqrt[3]{3}}\biggl(\sqrt[3]{\frac{x}{yz}}+\sqrt[3]{\frac{y}{zx}}+\sqrt[3]{\frac{z}{xy}}\biggl)\geq \frac{\sqrt[3]{3\sqrt[3]{xyz}}}{\sqrt[3]{3}}.3\sqrt[3]{\frac{1}{\sqrt[3]{xyz}}}\geq 3$
Therefore the proof is completed. Equality occurs for $x=y=z\Leftrightarrow a=b=c=1$

## August 21, 2011

### Problem 303 Van Khea

Filed under: Problem by Van Khea — KKKVVV @ 4:30 pm

If $a, b, c$ are positive real numbers then prove that

$\displaystyle \frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}\geq \frac{3\sqrt{3}}{4\sqrt{ab+bc+ca}}$

### Problem 302 Van Khea

Filed under: Problem by Van Khea — KKKVVV @ 3:34 pm

If $a, b, c$ are positive real numbers then prove that:

$\displaystyle \frac{a^2}{b\sqrt{c^2+3ab}}+\frac{b^2}{c\sqrt{a^2+3bc}}+\frac{c^2}{a\sqrt{b^2+3ca}}\geq \frac{3}{2}$
Solution
We have
$\displaystyle \frac{a^2}{b\sqrt{c^2+3ab}}+\frac{b^2}{c\sqrt{a^2+3bc}}+\frac{c^2}{a\sqrt{b^2+3ca}}$$\displaystyle =\frac{(a^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ab)^3}}+\frac{(b^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(bc)^3}}+\frac{(c^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ca)^3}}$
Because $\displaystyle \frac{3}{2}-\frac{1}{2}=1$ then we have
$\displaystyle \frac{(a^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ab)^3}}+\frac{(b^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(bc)^3}}+\frac{(c^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ca)^3}}$$\displaystyle \geq \frac{(a^2+b^2+c^2)^{\frac{3}{2}}}{\sqrt{3((ab)^3+(bc)^3+(ca)^3+a^2b^2c^2)}}$
Thus, it suffices to show that
$\displaystyle \frac{(a^2+b^2+c^2)^{\frac{3}{2}}}{\sqrt{3((ab)^3+(bc)^3+(ca)^3+a^2b^2c^2)}}\geq \frac{3}{2}$
$\displaystyle \Leftrightarrow 27((ab)^3+(bc)^3+(ca)^3+a^2b^2c^2)\leq 4(a^2+b^2+c^2)^3$
Letting $a^2=x; b^2=y; c^2=z$ yields
$\displaystyle 27((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)\leq 4(x+y+z)^3$
From $Cauchy-Schwarz$ inequality we have
$\displaystyle ((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)^2$$\leq (x^2y+y^2z+z^2x+xyz)(xy^2+yz^2+zx^2+xyz)$
So we need to prove that
$\displaystyle (x^2y+y^2z+z^2x+xyz)(xy^2+yz^2+zx^2+xyz)\leq \frac{16}{27^2}(x+y+z)^6$
Now we will show that for $x, y, z$ are positive real numbers then we have
$27(x^2y+y^2z+z^2x+xyz)\leq 4(x+y+z)^3$ ;$(1)$ and $27(xy^2+yz^2+zx^2+xyz)\leq 4(x+y+z)^3$ ; $(2)$
without loss of generality, suppose that $x=min(x, y, z)$. Sitting $y=x+u$ and $z=x+v; (u, v\geq 0)$ then:
$(1)\Leftrightarrow 9(u^2-uv+v^2)x+(u-2v)^2(4u+v)\geq 0$
$(2)\Leftrightarrow 9(u^2-uv+v^2)x+(2u-v)^2(u+4v)\geq 0$
which is obviously true. Equality occurs for $u=v=0\Leftrightarrow x=y=z$
$\displaystyle \Rightarrow ((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)^2$$\displaystyle \leq \frac{16}{27^2}(x+y+z)^6$
$\displaystyle \Leftrightarrow 27((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)\leq 4(x+y+z)^3$ is true.
Therefore the proof is completed. Equality occurs for $x=y=z\Leftrightarrow a=b=c$

### Problem 301 Van Khea

Filed under: Problem by Van Khea — KKKVVV @ 12:30 am

If $a, b, c$ are positive real numbers such that $abc=1$. Prove that

$\displaystyle \frac{a^{\frac{5}{8}}}{\sqrt{b+c}}+\frac{b^{\frac{5}{8}}}{\sqrt{c+a}}+\frac{c^{\frac{5}{8}}}{\sqrt{a+b}}\geq \frac{3}{\sqrt{2}}$
Proof
Let $a=x^4; b=y^4; c=z^4\Rightarrow xyz=1$ then we get:
$\displaystyle \frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\geq \frac{3}{\sqrt{2}}$
From $Cauchy-Scharz$ inequality we have
$\displaystyle (x^3+y^3+z^3)\biggl(\frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\biggl)$$\displaystyle \geq \biggl(\frac{x^{\frac{11}{4}}}{\sqrt[4]{y^4+z^4}}+\frac{y^{\frac{11}{4}}}{\sqrt[4]{z^4+x^4}}+\frac{z^{\frac{11}{4}}}{\sqrt[4]{x^4+y^4}}\biggl)^2$
Let $\displaystyle A=\frac{x^{\frac{11}{4}}}{\sqrt[4]{y^4+z^4}}+\frac{y^{\frac{11}{4}}}{\sqrt[4]{z^4+x^4}}+\frac{z^{\frac{11}{4}}}{\sqrt[4]{x^4+y^4}}$
We have $\displaystyle \frac{x^{\frac{11}{4}}}{\sqrt[4]{y^4+z^4}}=\frac{(x^3)^{\frac{5}{4}}}{\sqrt[4]{x^4y^4+z^4x^4}}$
Therefore we get $\displaystyle A=\frac{(x^3)^{\frac{5}{4}}}{\sqrt[4]{x^4y^4+z^4x^4}}+\frac{(y^3)^{\frac{5}{4}}}{\sqrt[4]{y^4z^4+x^4y^4}}+\frac{(z^3)^{\frac{5}{4}}}{\sqrt[4]{z^4x^4+y^4z^4}}$
Because $\displaystyle \frac{5}{4}-\frac{1}{4}=1$ so from $Problem$ $Van Khea$ we get
$\displaystyle A\geq \frac{(x^3+y^3+z^3)^{\frac{5}{4}}}{\sqrt[4]{2(x^4y^4+y^4z^4+z^4x^4)}}$$\displaystyle \Rightarrow A^2\geq \frac{(x^3+y^3+z^3)^{\frac{5}{2}}}{\sqrt{2(x^4y^4+y^4z^4+z^4x^4)}}$
$\displaystyle \Rightarrow (x^3+y^3+z^3)\biggl(\frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\biggl)$$\displaystyle \geq \frac{(x^3+y^3+z^3)^{\frac{5}{2}}}{\sqrt{2(x^4y^4+y^4z^4+z^4x^4)}}$
$\displaystyle \Rightarrow \frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\geq \frac{(x^3+y^3+z^3)^{\frac{3}{2}}}{\sqrt{2(x^4y^4+y^4z^4+z^4x^4)}}$
Now we will prove that if $x, y, z$ be positive real numbers then
$\displaystyle 3^{\frac{5}{3}}(x^4y^4+y^4z^4+z^4x^4)\leq (x^3+y^3+z^3)^{\frac{8}{3}}$
Let $\displaystyle u=\frac{x\sqrt[3]{3}}{\sqrt[3]{x^3+y^3+z^3}};v=\frac{y\sqrt[3]{3}}{\sqrt[3]{x^3+y^3+z^3}};w=\frac{z\sqrt[3]{3}}{\sqrt[3]{x^3+y^3+z^3}}$ then we just need to prove that with $u^3+v^3+w^3=3$ then $(uv)^4+(vw)^4+(wu)^4\leq 3$
From $AM-GM$ inequality we have
$\displaystyle uv\leq \frac{u^3+v^3+1}{3}=\frac{4-w^3}{3}$$\displaystyle \Rightarrow (uv)^4\leq \frac{4u^3v^3-u^3v^3w^3}{3}$
Therefore we get $\displaystyle (uv)^4+(vw)^4+(wu)^4\leq \frac{4((uv)^3+(vw)^3+(wu)^3)}{3}-u^3w^3w^3$
Thus, it suffices to show that: $4((uv)^3+(vw)^3+(wu)^3)-3u^3v^3w^3\leq 9$
Which is just the third degree $Schur's$ inequality
$4(rs+st+tr)(r+s+t)-3rst\leq (r+s+t)^3$
For $r=u^3; s=v^3; t=w^3$
Therefore we get $\displaystyle 3^{\frac{5}{3}}(x^4y^4+y^4z^4+z^4x^4)\leq (x^3+y^3+z^3)^{\frac{8}{3}}$
$\displaystyle \Rightarrow \frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\geq \frac{3^{\frac{5}{6}}(x^3+y^3+z^3)^{\frac{3}{2}}}{(x^3+y^3+z^3)^{\frac{4}{3}}\sqrt{2}}$$\displaystyle \geq \frac{3^{\frac{5}{6}}(x^3+y^3+z^3)^{\frac{1}{6}}}{\sqrt{2}}$
From $AM-GM$ inequality we have $x^3+y^3+z^3\geq 3xyz=3$
$\displaystyle \Rightarrow \frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\geq \frac{3}{\sqrt{2}}$
Therefore the proof is completed. Equality occurs for $x=y=z=1\Leftrightarrow a=b=c=1$

## August 19, 2011

### Problem 299 Van Khea

Filed under: Problem by Van Khea — KKKVVV @ 5:51 am

Let $a, b, c$ be positive real numbers such that $a^4+b^4+c^4=3$. Prove that

$\displaystyle \frac{a^3}{\sqrt{b^4+c^4}}+\frac{b^3}{\sqrt{c^4+a^4}}+\frac{c^3}{\sqrt{a^4+b^4}}\geq \frac{3}{\sqrt{2}}$

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