$3 \ge \frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}=\frac{(xy)^2+(yz)^2+(zx)^2}{xyz} \ge \frac{xyz(x+y+z)}{xyz}=x+y+z \Rightarrow x+y+z \le \ 3$; $(o)$ So, $\frac{x^2}{y^3}+\frac{y^2}{z^3}+\frac{z^2}{x^3}=\frac{(\frac{x}{y})^3}{x}+\frac{(\frac{y}{z})^3}{y}+\frac{(\frac{z}{x})^3}{z} \ge (*) \frac{(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})^3}{3(x+y+z)} \ge (o) \frac{(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})^3}{9} \ge (**) \frac{x}{y}+\frac{y}{z}+\frac{z}{x}$, q.e.d., where: (*) is Holder's inequality; (**) is $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \ge \ 3$, which is AM-GM. Equality hold if and only if $x=y=z=1$.

Or: So, $\frac{x^2}{y^3}+\frac{y^2}{z^3}+\frac{z^2}{x^3}=\frac{(\frac{x}{y})^2}{y}+\frac{(\frac{y}{z})^2}{z}+\frac{(\frac{z}{x})^2}{x} \ge (*)\frac{(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})^2}{x+y+z} \ge (o)\ge \frac{(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})^2}{3} \ge (**)\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$, q.e.d, where: $(*)$ is T2's Lemma: $(**)$ is $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \ge \ 3$ (AM-GM).

by AM.GM inequality : $$\frac{x^2}{y^3}+y\geq 2\frac{x}{y}$$and similary.Thus $$\sum \frac{x^2}{y^3}+x+y+z\geq 2\sum \frac{x}{y}$$so we need to prove $$ 2\sum \frac{x}{y}-\sum x \geq \sum \frac{x}{y}$$ie $$ \sum \frac{x}{y}\geq \sum x$$by CS :$$\sum \frac{x}{y}\geq \frac{(x+y+z)^2}{xy+yz+xz} \geq x+y+z$$ie $$x+y+z\geq xy+yz+xz$$but $$xy+yz+xz\le \frac{((x+y+z)^2}{3}\le x+y+z$$ie $$x+y+z\le 3$$ture by condition

nimueh wrote: For all positive real numbers $x,y,z$ satisfying the inequality $\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\leq 3$, prove that $\frac{x^2}{y^3}+\frac{y^2}{z^3}+\frac{z^2}{x^3}\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ Proof of Zhangyanzong :

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nimueh wrote: For all positive real numbers $x,y,z$ satisfying the inequality $\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\leq 3$, prove that $\frac{x^2}{y^3}+\frac{y^2}{z^3}+\frac{z^2}{x^3}\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ The following inequality is also true

nimueh wrote: For all positive real numbers $x,y,z$ satisfying the inequality $\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\leq 3$, prove that $\frac{x^2}{y^3}+\frac{y^2}{z^3}+\frac{z^2}{x^3}\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ $C-S$ gives $(\sum{\dfrac{x^2}{y^3}})(\sum{x}) \ge (\sum{\dfrac{x}{y}})^2$. So it suffices to prove that $\sum{\dfrac{x}{y}} \ge \sum{x}$. It is $3 \ge \sum{\dfrac{xy}{z}} \ge x+y+z$ $\Leftrightarrow$ $x+y+z \le 3 \le \dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}$. Hence, the result. Equality iff $x=y=z=1$

nimueh wrote: For all positive real numbers $x,y,z$ satisfying the inequality $\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\leq 3$, prove that $\frac{x^2}{y^3}+\frac{y^2}{z^3}+\frac{z^2}{x^3}\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ What is EGMO for ?

knm2608 wrote: nimueh wrote: For all positive real numbers $x,y,z$ satisfying the inequality $\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\leq 3$, prove that $\frac{x^2}{y^3}+\frac{y^2}{z^3}+\frac{z^2}{x^3}\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ What is EGMO for ? I think it's math olympiad for girls or somthing like that!

knm2608 wrote: nimueh wrote: For all positive real numbers $x,y,z$ satisfying the inequality $\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\leq 3$, prove that $\frac{x^2}{y^3}+\frac{y^2}{z^3}+\frac{z^2}{x^3}\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ What is EGMO for ? European Girl Mathematical Olympiad.

OK thanks.

Same as pure solution

nimueh wrote: For all positive real numbers $x,y,z$ satisfying the inequality $\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\leq 3$, prove that $\frac{x^2}{y^3}+\frac{y^2}{z^3}+\frac{z^2}{x^3}\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ We need to prove that $$\sum_{cyc}\frac{x^2}{y^3}\sum_{cyc}\frac{xy}{z}\geq3\sum_{cyc}\frac{x}{y}$$or $$\sum_{cyc}(x^7z^3y^2+x^7z^5+x^5y^5z^2-3x^5z^4y^3)\geq0,$$which is AM-GM.

@below nicw

We need to prove that $$\frac{x^2}{y^3}+\frac{y^2}{z^3}+\frac{z^2}{x^3}\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}$$or $$3\left(\frac{x^2}{y^3}+\frac{y^2}{z^3}+\frac{z^2}{x^3}\right)\geq3\left( \frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right).$$Now, $$3\left(\frac{x^2}{y^3}+\frac{y^2}{z^3}+\frac{z^2}{x^3}\right)\geq\sum_{cyc}\frac{x^2}{y^3}\sum_{cyc}\frac{xy}{z}.$$Thus, it's enough to prove that: $$\sum_{cyc}\frac{x^2}{y^3}\sum_{cyc}\frac{xy}{z}\geq3\sum_{cyc}\frac{x}{y}$$

nimueh wrote: For all positive real numbers $x,y,z$ satisfying the inequality $$\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\leq 3,$$prove that $$\frac{x^2}{y^3}+\frac{y^2}{z^3}+\frac{z^2}{x^3}\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}.$$

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Ok, my solution seems unnecessary, say $a=\frac{xy}{z}$ etc. We need to prove that while $a+b+c\leq 3$, $$S=\sum_{cyc}\frac{c}{b\sqrt{ab}}\geq \sum_{cyc}\sqrt{\frac{c}{b}}$$Because $a+b+c\leq 3$, $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3$, so$$3S \geq S\cdot \left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\geq \left(\sum_{cyc}\sqrt{\frac{c}{b}}\right)^2 \geq 3\cdot \left(\sum_{cyc}\sqrt{\frac{c}{b}}\right)$$because $\sum_{cyc}\sqrt{\frac{c}{b}}\geq 3$ by AM-GM which finishes the proof.