Derive the conjugate of the negative geometric mean $$f(x) = - \left( \prod_{i=1}^n x_i \right)^{\frac1n}.$$
Here, $x = (x_1,\dots,x_n)$. In the solution manual here it is stated that
Next assume that $y \preceq 0$, but $(\prod_i (-y_i))^{1/n}<1/n$. We choose $x_i=-t/y_i$ and obtain $$x^Ty-f(x)=-tn-t\left(\prod_i(-1/y_i) \right)^{1/n}\to\infty$$ as $t\to\infty$.
But the last expression is $-tn-tz$ for some $z>0$ so it is always negative. It approaches $-\infty$ as $t\to\infty$. What am I missing?
Yes, your correction in the comment is correct. Here is a mild elaboration: \begin{align}x^Ty - f(x) &= \sum_{i=1}^n\frac {-t}{y_i} y_i -\left(- \left( \prod_{j=1}^n \frac{-t}{y_j}\right)^{1/n} \right) &(\text{def. of $f$ and $x$})\\&= -tn +t \left( \prod_{j=1}^n \frac{-1}{y_j}\right)^{1/n} &(\text{simplifying})\\&= t\left(-n + \frac{1}{( \prod_{j=1}^n(-y_j))^{1/n}} \right) &(\text{rearranging to use givens})\end{align} and the last quantity in the bracket is positive by the assumption that $$ \left( \prod_{j=1}^n(-y_j)\right)^{1/n} < \frac1n \iff \frac{1}{( \prod_{j=1}^n(-y_j))^{1/n}} >n,$$ hence the limit as $t\to\infty$ is $\infty$.