let $x,y,z>0$,show that $$\sqrt{\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)}+1\ge 2\sqrt[3]{\dfrac{(x^2+yz)(y^2+xz)(z^2+xy)}{x^2y^2z^2}}$$

My try: $$\Longleftrightarrow \left(\sqrt{\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)}+1\right)^3\ge \dfrac{8(x^2+yz)(y^2+xz)(z^2+xy)}{x^2y^2z^2}$$ let $$a=\dfrac{x}{y},b=\dfrac{y}{z},c=\dfrac{z}{x}$$ $$\Longleftrightarrow \left(\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)(a+b+c)}+1\right)^3\ge 8\left(1+\dfrac{c}{a}\right)\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)$$

then I can't ,so I think this inequality maybe have other nice methods,Thank you


1 Answer 1


As the inequality is homogeneous, we can normalise by $xyz=1$. Then we have:

$$\sqrt{\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)}+1\ge 2\sqrt[3]{\left(x^2+\frac{1}{x}\right)\left(y^2+\frac{1}{y}\right)\left(z^2+\frac{1}{z}\right)}$$

$$\sqrt{3+\sum_{cyc} \left(x^3 + \frac{1}{x^3}\right)}+1\ge 2\sqrt[3]{2 + \sum_{cyc} \left(x^3 + \frac{1}{x^3}\right)}$$

Let $\displaystyle a = \sum_{cyc} \left(x^3 + \frac{1}{x^3}\right)\ge 6$. Then the inequality is reduced to $$f(a) = \sqrt{3+a} +1 - 2 \sqrt[3]{2+a} \ge 0$$ which is easy to do as $f(6)=0$ and $f'(a) > 0 $ for $a > 6$.

Addendum: alternate way to show $\sqrt{3+a} +1 \ge 2 \sqrt[3]{2+a}$ would be to cube, group terms and then square, to get the equivalent $(a+2)(a-6)^2 \ge 0$, which is obvious for $a \ge 6$.

  • $\begingroup$ Can you explain how did you figure out that $xyz=1$? $\endgroup$
    – Ovi
    Mar 3, 2014 at 11:10
  • $\begingroup$ @Ovi As the inequality is homogeneous, i.e. scaling all variables by the same factor will not change the inequality. Hence we can select such a factor so that $xyz=1$. If it were more convenient, we could have instead set $x+y+z=2$ or something else like $\max(x,y,z)=3$. $\endgroup$
    – Macavity
    Mar 3, 2014 at 11:40

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