# 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 $x=y=z=1\Leftrightarrow a=b=c=1$