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Let $$z:=\frac{a+b+c}{\sqrt[3]{abc}}.$$ Prove that for $n\leq 3$, $$z+\frac{3n}{z}\geq 3+n.$$

This was written here and I couldn't understand it. Proving inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$

Can anyone explain?

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  • $\begingroup$ If $z>0$ $z+\frac{3n}{z}\ge 3+n \implies z^2-(3+n)z+3n \ge 0$ As $\Delta= (3+n)^2-12n=(n-3)^2 \le 0 \implies n=3$ $\endgroup$
    – Z Ahmed
    Oct 12, 2021 at 7:19

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I will use the assumptions from your linked post that $a,b,c\in\mathbb{R}^+$. We have by AM-GM that $$\frac{a+b+c}{3}\geq\sqrt[3]{abc}$$ $$\frac{a+b+c}{\sqrt[3]{abc}}\geq 3$$ $$z\geq 3$$

Note that we have $$z+\frac{3n}{z}\geq 3+n$$ $$\iff z-3\geq \frac{z-3}{z}n$$ $$\iff \frac{(z-3)(z-n)}{z}\geq 0$$ We can easily see that $z-3$, $z-n$, and $z$ will all be nonnegative. Hence it follows that the entire expression will be nonnegative.

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