$$\begin{array}{ll} \text{minimize} & f(x) + g(y)\\ \text{subject to} & xy \ge a\\ & x \ge 0\\ & y \ge 0\end{array}$$
where both $f$ and $g$ are convex quadratic functions and $a > 0$. The feasible region is convex and, since $a > 0$, also LMI-representable
$$\{ (x,y) \in \mathbb R^2 : x \geq 0 \land y \geq 0 \land x y \geq a \} = \left\{ (x,y) \in \mathbb R^2 : \begin{bmatrix} x & \sqrt{a}\\ \sqrt{a} & y\end{bmatrix} \succeq \mathrm O_2 \right\}$$
Hence, the original optimization problem can be rewritten as follows
$$\begin{array}{ll} \text{minimize} & f(x) + g(y)\\ \text{subject to} & \begin{bmatrix} x & \sqrt{a}\\ \sqrt{a} & y\end{bmatrix} \succeq \mathrm O_2\end{array}$$
Introducing optimization variables $s, t \in \mathbb R$, we rewrite the optimization problem in epigraph form
$$\begin{array}{ll} \text{minimize} & s + t\\ \text{subject to} & f(x) \leq s\\ & g(y) \leq t\\ & \begin{bmatrix} x & \sqrt{a}\\ \sqrt{a} & y\end{bmatrix} \succeq \mathrm O_2\end{array}$$
Let $f$ and $g$ be
$$f (x) := f_0 + f_1 x + f_2 x^2 \qquad\qquad\qquad g (y) := g_0 + g_1 y + g_2 y^2$$
where $f_2, g_2 > 0$ (to ensure convexity). Inequality constraints $f(x) \leq s$ and $g(y) \leq t$ can be written in LMI form, as follows
$$\begin{bmatrix} 1 & \sqrt{f_2} \, x\\ \sqrt{f_2} \, x & s - f_0 - f_1 x\end{bmatrix} \succeq \mathrm O_2$$
$$\begin{bmatrix} 1 & \sqrt{g_2} \, y\\ \sqrt{g_2} \, y & t - g_0 - g_1 y\end{bmatrix} \succeq \mathrm O_2$$
These LMIs introduce inequalities $s - f_0 - f_1 x \geq 0$ and $t - g_0 - g_1 y \geq 0$, which are redundant. Note that lines $s = f_0 + f_1 x$ and $t = g_0 + g_1 y$ are tangent to the graphs of $f$ and $g$, respectively.
Hence, we obtain a semidefinite program (SDP) in variables $x, y, s, t \in \mathbb R$
$$\begin{array}{lc} \\ \text{minimize} & s + t\\\\ \text{subject to} & \begin{bmatrix} 1 & \sqrt{f_2} \, x & & & \\ \sqrt{f_2} \, x & s - f_0 - f_1 x & & & \\ & & 1 & \sqrt{g_2} \, y & \\ & & \sqrt{g_2} \, y & t - g_0 - g_1 y & \\ & & & & x & \sqrt{a}\\ & & & & \sqrt{a} & y\end{bmatrix} \succeq \mathrm O_6\\\\\end{array}$$
which can be solved numerically using any SDP solver.
A problem with integer variables is non-convex pretty much by definition of convexity (a set consisting of isolated points like $\{0,1\}$ is not convex). However, by a slight abuse of language, "mixed-integer convex problem" is used to describe mixed-integer problems where the continuous relaxation is convex. They are already hard because they contain the NP-hard class of mixed-integer linear problems. Like you say, a solver such as MOSEK, and a tool such as CVX can often be used to model and solve such type of problems.
Your problem is not of that form ie. it is even harder because the nonlinear constraints 1 and 2 are non-convex. (Constraint 3 is just a roundabout way to write that $d$ is binary, which you would communicate to the solver more directly anyway). Unless there is some special structure to exploit then you may need a general MINLP solver (see for example list of MINLP solvers in https://en.wikipedia.org/wiki/List_of_optimization_software)
Best Answer
You're in luck: this is a geometric program. Therefore, the change of variables $$x\rightarrow e^{u}, ~~ y\rightarrow e^{v}, ~~ z\rightarrow e^{w}$$ will get us close, and a few logarithms will get us the rest of the way there. Substitution yields \begin{array}{ll} \text{minimize} & e^{u-v} \\ \text{subject to} & 2 \leq e^u \leq 3 \\ & e^{2u} + e^{v-w} \leq e^{v/2} \\ & e^{u-v}=e^{2w} \end{array} Now let's take some logarithms: \begin{array}{ll} \text{minimize} & u-v \\ \text{subject to} & \log 2 \leq u \leq \log 3 \\ & \log \left( e^{2u} + e^{v-w} \right) \leq v/2 \\ & u-v=2w \end{array} And there you have it: a convex optimization problem. The "log-sum-exp" expression is indeed a smooth convex function; you'll have to take that on faith or prove it to yourself. Recovering $x,y,z$ once you have solved for $u,v,w$ just requires three exponentials.
Some notes:
We could skip a few of the logarithms and still have a convex problem: the objective, the $e^u\leq 3$ constraint, the nonlinear inequality. But in practice we keep those logarithms in there; the problems are better behaved numerically that way. (I have no theoretical support for this; someone else might, though). The other logarithms must be taken, however.
Strictly speaking, the convex version of this model is equivalent only if we assume strict positivity on $x,y,z$. Yes, you can handle the case of zero variables with more technical machinery; but fortunately we don't need that here. After all, $x\geq 2$ is guaranteed from the constraints; and $y$ cannot be zero unless the entire problem is infeasible, because of the $x/y$ objective. And therefore, from the $x/y=z^2$ constraint, we know that $z$ is nonzero as well.
If we wish we can eliminate $z$ and $w$ altogether, by taking advantage of the equality constraint, to yield
\begin{array}{lllll} \text{minimize} & x/y & & \text{minimize} & u-v \\ \text{subject to} & 2 \leq x \leq 3 & & \text{subject to} & \log 2 \leq u \leq \log 3 \\ & x^2 + y^{3/2} x^{-1/2} \leq y^{1/2} && & \log \left( e^{2u} + e^{(3v-u)/2} \right) \leq v/2 \\ \end{array}
I cannot emphasize the following enough, but I am going to try:
The title of this question suggests the author is making this assumption. Perhaps that's because this problem was assigned as a homework or test question, in which case it was known in advance by the question designer to be a geometric program. But geometric programs are the exception, not the rule. I'm unaware of any other general, non-manufactured class of nonconvex problems that can be made convex with just a change of variables.