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If $x,y,z$ are unequal positive quatities then prove that , $(1+x^3)(1+y^3)(1+z^3)>(1+xyz)^3$

Soubhik
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    Hello - please tell the forum what you have tried, whether you can think of an approach, whether you have seen a similar problem etc. – Hans Engler Jun 14 '14 at 14:02

2 Answers2

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lemma: if $a,b,c,x,y,z,p,q,r\ge 0$ then we have $$(a^3+b^3+c^3)(x^3+y^3+z^3)(p^3+q^3+r^3)\ge (axp+byq+czr)^3$$

poof: note and Use AM-GM inequality \begin{align*}3&=\sum_{cyc}\dfrac{a^3}{a^3+b^3+c^3}+\sum_{cyc}\dfrac{x^3}{x^3+y^3+z^3}+\sum_{cyc}\dfrac{p^3}{p^3+q^3+r^3}\\ &\ge\sum_{cyc}\dfrac{3axp}{\sqrt[3]{(a^3+b^3+c^3)(x^3+y^3+z^3)(p^3+q^3+r^3)}}\\ \end{align*} so

$$axp+byq+czr\le \sqrt[3]{(a^3+b^3+c^3)(x^3+y^3+z^3)(p^3+q^3+r^3)}$$

math110
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Hint: Factor it out and cancel like terms from both sides. Then, rearrange so that the right side equals $0$.

Aidan F. Pierce
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