AlperenINAN wrote: Let $x,y,z$ are positive real numbers that satisfy at least one of the inequalities of $2xy>1$, $yz>1$. Find the least possible value of $xy^3z^2+\frac{4z}{x}-8yz-\frac{4}{yz}$ . Maybe,$$ - 8$$

For fixed $x$ and $y,$ denoting $f(x) = xy^3z^2 + \frac{4z}{x}$, we have that $f(x)$ decreases in $\left(0, \frac{2}{y\sqrt{yz}}\right]$ and increases in $\left[\frac{2}{y\sqrt{yz}}, \infty\right)$. Suppose firstly that $yz>1$. Then we have no restrictions on $x$, so the main expression is at least $4(yz)^{3/2} - 8yz - \frac{4}{yz}$ and so with $t=\sqrt{yz} > 1$ we aim to minimize $t^3 - 2t^2 - \frac{1}{t^2}$. The best we can show is that $t^3 - 2t^2 - \frac{1}{t^2} > -2$, equivalent to $t>1$ (and the limit as $t\to 1$ is $(-2)$). The corresponding infimum (but not minimum, hmm?) of the expression is $(-8)$. Now suppose that $x > \frac{1}{2y}$. Then if $\frac{1}{2y} \leq \frac{2}{y\sqrt{yz}}$, we get $yz \leq 16$, the expression is minimized for $x$ as above. So suppose $yz > 16$. Then the expression increases in $\left(\frac{1}{2y}, \infty\right)$ and hence it exceeds $\frac{y^2z^2}{2} - \frac{4}{yz} = \frac{s^2}{2} - \frac{4}{s}$ where $s>16$ and the latter has positive derivative, so exceeds $127.75$.