INEQUALITY'S BLOG

July 17, 2011

Problem 239 (vankhea)

Filed under: Problem by Van Khea — KKKVVV @ 6:17 am

Prove that for a, b, c, p, q>0 we have: \displaystyle \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}\geq \frac{3}{(abc)^{\frac{4}{9}}.\sqrt[3]{p+q}}

July 16, 2011

Problem 238 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 11:13 am

Prove that for a, b, c>0 we have:

\displaystyle \frac{1}{4a}+\frac{1}{4b}+\frac{1}{4c}+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\displaystyle \geq \frac{9}{4}\biggl(\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\biggl)

July 14, 2011

Problem 237 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 7:14 am

Let \displaystyle a, b, c\geq 0\&  \frac{1}{3a+1}+\frac{1}{3a+1}+\frac{1}{3a+1}=3. Prove that:

\displaystyle \frac{1}{3a+2}+\frac{1}{3b+2}+\frac{1}{3c+2}\geq \frac{1}{2}\biggl(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\biggl)

July 13, 2011

Problem 236 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 4:20 am

Let a, b, c>0\& a^3+b^3+c^3=3. Prove that: (ab)^4+(bc)^4+(ca)^4+2(ab)^3+2(bc)^3+2(ca)^3\leq 9
Proof
We have \displaystyle (ab)^4+(bc)^4+(ca)^4\leq 9-2((ab)^3+(bc)^3+(ca)^3)\leq (a^3+b^3+c^3)^2-2((ab)^3+(bc)^3+(ca)^3)

\Rightarrow (ab)^4+(bc)^4+(ca)^4\leq a^6+b^6+c^6

Let x=a^3; y=b^3; z=c^3\Rightarrow x+y+z=3 then we have \displaystyle (xy)^{\frac{4}{3}}+(yz)^{\frac{4}{3}}+(zx)^{\frac{4}{3}}\leq x^2+y^2+z^2

We have \displaystyle (xy)^{\frac{4}{3}}=(x^2y)^{\frac{1}{3}}(xy^2)^{\frac{1}{3}}(xy)^{\frac{1}{3}}

\displaystyle \Rightarrow (xy)^{\frac{4}{3}}+(yz)^{\frac{4}{3}}+(zx)^{\frac{4}{3}}\displaystyle =(x^2y)^{\frac{1}{3}}(xy^2)^{\frac{1}{3}}(xy)^{\frac{1}{3}}+(y^2z)^{\frac{1}{3}}(yz^2)^{\frac{1}{3}}(yz)^{\frac{1}{3}}+(z^2x)^{\frac{1}{3}}(zx^2)^{\frac{1}{3}}(zx)^{\frac{1}{3}}

From Holder’s inequality we have

\displaystyle ((x^2y)^{\frac{1}{3}}(xy^2)^{\frac{1}{3}}(xy)^{\frac{1}{3}}+(y^2z)^{\frac{1}{3}}(yz^2)^{\frac{1}{3}}(yz)^{\frac{1}{3}}+(z^2x)^{\frac{1}{3}}(zx^2)^{\frac{1}{3}}(zx)^{\frac{1}{3}})^3

\displaystyle \leq (x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)(xy+yz+zx)

\displaystyle \Leftrightarrow ((xy)^{\frac{4}{3}}+(yz)^{\frac{4}{3}}+(zx)^{\frac{4}{3}})^3\displaystyle \leq (x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)(xy+yz+zx)

But we have

\displaystyle x^3+y^3+z^3\geq \frac{1}{3}(x+y+z)(x^2+y^2+z^2)\geq xy+yz+zx

\displaystyle \Rightarrow ((xy)^{\frac{4}{3}}+(yz)^{\frac{4}{3}}+(zx)^{\frac{4}{3}})^3\displaystyle \leq (x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)(x^3+y^3+z^3)

From AM-GM inequality we have

\displaystyle (x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)(x^3+y^3+z^3)\displaystyle \leq \biggl(\frac{x^2y+y^2z+z^2x+xy^2+yz^2+zx^2+x^3+y^3+z^3}{3}\biggl)^3\displaystyle \leq \biggl(\frac{(x+y+z)(x^2+y^2+z^2)}{3}\biggl)^3=(x^2+y^2+z^2)^3

\displaystyle \Rightarrow ((xy)^{\frac{4}{3}}+(yz)^{\frac{4}{3}}+(zx)^{\frac{4}{3}})^3\leq (x^2+y^2+z^2)^3

\displaystyle \Rightarrow (xy)^{\frac{4}{3}}+(yz)^{\frac{4}{3}}+(zx)^{\frac{4}{3}}\leq x^2+y^2+z^2

Therefore the proof is completed. Equality hold if x=y=z=1\Leftrightarrow a=b=c=1

Problem 235 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 2:43 am

Let a, b, c\geq 0. Prove that: (a^2+b^2+c^2)^3\geq 8\biggl[(ab)^3+(bc)^3+(ca)^3\biggl]

Problem 234 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 1:02 am

Let a, b, c\geq 0. Prove that: \displaystyle (a^6+b^6+c^6)^7\geq 81\biggl[(ab)^7+(bc)^7+(ca)^7\biggl]^3

Proof

Let x=a^6; y=b^6; z=c^6 then we need to prove that:

\displaystyle (x+y+z)^7\geq 81\biggl[(xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}}\biggl]^3

We have \displaystyle (xy)^{\frac{7}{6}}=(x^3)^{\frac{1}{18}}(y^3)^{\frac{1}{18}}(xy^2)^{\frac{1}{9}}(x^2y)^{\frac{1}{9}}(xy)^{\frac{2}{3}}

\displaystyle \Rightarrow (xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}}

\displaystyle =(x^3)^{\frac{1}{18}}(y^3)^{\frac{1}{18}}(xy^2)^{\frac{1}{9}}(x^2y)^{\frac{1}{9}}(xy)^{\frac{2}{3}}+(y^3)^{\frac{1}{18}}(z^3)^{\frac{1}{18}}(yz^2)^{\frac{1}{9}}(y^2z)^{\frac{1}{9}}(yz)^{\frac{2}{3}}\displaystyle +(z^3)^{\frac{1}{18}}(x^3)^{\frac{1}{18}}(zx^2)^{\frac{1}{9}}(z^2x)^{\frac{1}{9}}(zx)^{\frac{2}{3}}

Because \displaystyle \frac{1}{18}+\frac{1}{18}+\frac{1}{9}+\frac{1}{9}+\frac{2}{3}=1 then from Holder's inequality we get:

\displaystyle (xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}}\displaystyle \leq (x^3+y^3+z^3)^{\frac{2}{18}}(x^2y+y^2z+z^2x)^{\frac{1}{9}}(xy^2+yz^2+zx^2)^{\frac{1}{9}}(xy+yz+zx)^{\frac{2}{3}}

\displaystyle \Leftrightarrow (xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}}\displaystyle \leq ((x^3+y^3+z^3)(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2))^{\frac{1}{9}}(xy+yz+zx)^{\frac{2}{3}}

From AM-GM inequality we have

\displaystyle (x^3+y^3+z^3)(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\leq \biggl(\frac{x^3+y^3+z^3+x^2y+y^2z+z^2x+xy^2+yz^2+zx^2}{3}\biggl)^3

\displaystyle \Leftrightarrow (x^3+y^3+z^3)(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\leq \biggl(\frac{(x+y+z)(x^2+y^2+z^2)}{3}\biggl)^3

\displaystyle \Rightarrow (xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}}\leq \biggl(\frac{(x+y+z)(x^2+y^2+z^2)}{3}\biggl)^{\frac{1}{3}}\displaystyle (xy+yz+zx)^{\frac{2}{3}}

\displaystyle \Leftrightarrow (xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}}\leq \frac{(x+y+z)^{\frac{1}{3}}((x^2+y^2+z^2)(xy+yz+zx)^2)^{\frac{1}{3}}}{\sqrt[3]{3}}

From AM-GM inequality we have

\displaystyle (x^2+y^2+z^2)(xy+yz+zx)^2\leq \biggl(\frac{x^2+y^2+z^2+2(xy+yz+zx)}{3}\biggl)^3

\displaystyle \Leftrightarrow (x^2+y^2+z^2)(xy+yz+zx)^2\leq \frac{(x+y+z)^6}{27}

\displaystyle \Rightarrow (xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}}\leq \frac{(x+y+z)^{\frac{1}{3}}(x+y+z)^2}{3\sqrt[3]{3}}\displaystyle =\frac{(x+y+z)^{\frac{7}{3}}}{3\sqrt[3]{3}}

\displaystyle \Rightarrow (x+y+z)^7\geq 81((xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}})^3

Therefore the proof is completed. Equality holds for x=y=z\Leftrightarrow a=b=c

July 12, 2011

Problem 233 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 12:10 am

Let a, b, c be the lengths of the sides of a trangle. Prove that: \displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\geq 6+\frac{1}{2}\biggl[\frac{(a-b)^2}{c^2}+\frac{(b-c)^2}{a^2}+\frac{(c-a)^2}{b^2}\biggl]

Problem 232 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 12:03 am

Let a, b, c be the lengths of the sides of a trangle such that abc=1 . Prove that: \displaystyle (a-1)^2+(b-1)^2+(c-1)^2\geq \frac{1}{2}((ab-1)^2+(bc-1)^2+(ca-1)^2)

July 11, 2011

Problem 231 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 11:56 pm

Let a, b, c>0\& a+b+c=1. Prove that: \displaystyle \frac{1}{a^2(1-b)^2}+\frac{1}{b^2(1-b)^2}+\frac{1}{c^2(1-c)^2}\geq \frac{9}{4abc}

Problem 230 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 11:52 pm

Let a, b, c>0\& a^2+b^2+c^2=1. Prove that: \displaystyle \frac{1}{a^2-b^2c^2}+\frac{1}{b^2-c^2a^2}+\frac{1}{c^2-a^2b^2}\leq \frac{27}{4}

Problem 229 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 11:46 pm

Prove that for any positive real numbers a, b, c we have: \displaystyle \frac{1}{a^3(b^3+abc)}+\frac{1}{b^3(c^3+abc)}+\frac{1}{c^3(a^3+abc)}\geq \frac{3}{2a^2b^2c^2}

Problem 228 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 11:39 pm

Let a, b, c>0\& a^2+b^2+c^2=3. Prove that: \displaystyle \frac{a^2}{1+b^3}+\frac{b^2}{1+c^3}+\frac{c^2}{1+a^3}\geq \frac{3}{2}

Problem 227 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 11:34 pm

Let a, b, c>0\& abc=1. Prove that: \displaystyle \frac{a^{\frac{2}{3}}}{\sqrt{1+b}}+\frac{b^{\frac{2}{3}}}{\sqrt{1+c}}+\frac{c^{\frac{2}{3}}}{\sqrt{1+a}}\geq \frac{3}{\sqrt{2}}

Problem 226 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 11:29 pm

Let \displaystyle a, b, c>0\& a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}=3. Prove that: \displaystyle \frac{a^{\frac{2}{3}}}{\sqrt{b+c}}+\frac{b^{\frac{2}{3}}}{\sqrt{c+a}}+\frac{c^{\frac{2}{3}}}{\sqrt{a+b}}\geq \frac{3}{\sqrt{2}}

Problem 225 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 11:24 pm

a, b, c>0\& abc=1. Prove that: \displaystyle \frac{a^3}{b^2+c^3}+\frac{b^3}{c^2+a^3}+\frac{c^3}{a^2+b^3}\geq \frac{3}{2}

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