Ferum_2710 wrote: Determine the real numbers $x$, $y$, $z > 0$ for which $xyz \leq \min\left\{\frac{y}{x-1}, \frac{z}{y-1}, \frac{x}{z-1}\right\}$ Uhhh ? There are infinitely many such $x,y,z$ (for example any $x,y,z\in(1,\frac 43)$)

pco wrote: Ferum_2710 wrote: Determine the real numbers $x$, $y$, $z > 0$ for which $xyz \leq \min\left\{\frac{y}{x-1}, \frac{z}{y-1}, \frac{x}{z-1}\right\}$ Uhhh ? There are infinitely many such $x,y,z$ (for example any $x,y,z\in(1,\frac 43)$) my bad sorry

Ferum_2710 wrote: Determine the real numbers $x$, $y$, $z > 0$ for which $xyz \leq \min\left\{4(x - \frac{1}{y}), 4(y - \frac{1}{z}), 4(z - \frac{1}{x})\right\}$ Let $xy=a,yz=b,zx=c$. Then the given condition easily rewrites as $\min \{a(4-b),b(4-c),c(4-a) \} \geq 4$. Therefore, by the AM-GM inequality, $64 \leq a(4-b) \cdot b(4-c) \cdot c(4-a)=a(4-a) \cdot b(4-b) \cdot c(4-c) \leq 4^3=64,$ and so we must have equality everywhere. Thus, $a=b=c=2$, that is $x=y=z=\sqrt{2}$.

Note that $xyz \leq 4\left(x - \frac{1}{y}\right)$ requires $4\sqrt{xz} \leq xyz + \frac{4}{y} \leq 4x$ by AM-GM, thus $z\leq x$. Analogously from the other two we get $x \leq y$ and $y \leq z$, so necessarily $x=y=z$. Now the requirement holds if and only if $x^3 \leq 4\left(x - \frac{1}{x}\right)$, i.e. $(x^2 - 2)^2 \leq 0$, i.e. $x=\sqrt{2}$, so $x=y=z=\sqrt{2}$ is the only solution.