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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