INEQUALITY'S BLOG

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

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