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Given $a, b, c, t > 0$, let function $f : \Bbb R_{>0} \times \Bbb R_{>0} \to \Bbb R_{>0}$ be defined by

$$ f(x,y) := b \exp{\left(\frac{a x} {t x- c }\right)} \left[\exp{\left( \frac{ay}{ty - c} \right)} - 1 \right] $$

where $t x- c > 0$ and $ty - c >0$. I want to determine whether function $f$ is convex.

I tried the get the second derivative but the derivatives are too complicated to show the convexity. The exponential function is convex and if the function inside the $\exp$ if convex then $\exp$ (the function) is convex. In my problem, the function inside the exp is a linear fractional which is quasiconvex. So, I am not sure which operation preserves convexity I can use. Suggestions would be welcome.

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  • $\begingroup$ This function is convex if and only if $g:x\to \exp\left( \frac{ax}{tx-c} \right)$ is convex. $\endgroup$
    – P. Quinton
    Commented Aug 28, 2021 at 6:44

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If $g:x\to \exp(\frac{ax}{tx-c})$ is convex then $f$ is convex.

From wolfram alpha we get that the second derivative of $g$ is $g''(x)=\frac{ac g(x) (ac+2t(tx-c))}{(c-tx)^4}$ which is positive. Therefore $f$ is convex.

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