= Solution
The minimum exceeds $t$ exactly when both times exceed $t$. By <independence>,
$$
\mathbb P(X>t)=F_A(t)F_B(t).
$$
Differentiating its <survivor function> gives
$$
\boxed{f_X(t)=-\frac{d}{dt}[F_A(t)F_B(t)]
=f_A(t)F_B(t)+F_A(t)f_B(t).}
$$
The <tail integral formula for moments> then gives
$$
\boxed{\mathbb E X=\int_0^\infty F_A(t)F_B(t)\,dt.}
$$
Equivalently, <integration by parts> in $\int_0^R t f_X(t)\,dt$ gives $-RF_A(R)F_B(R)+\int_0^R F_A(t)F_B(t)\,dt$. The assumed $tF_B(t)\to0$ makes the boundary term vanish, since $0\leq F_A\leq1$.
The stated tail condition does not itself guarantee a finite mean. For example, the proper <survivor functions>
$$
F_B(t)=\frac1{(1+t)\log(e+t)},\qquad
F_A(t)=\frac1{\log(e+\log(1+t))}
$$
satisfy $tF_B(t)\to0$, but their product behaves as $1/[t\log t\log\log t]$, whose integral diverges. Thus the displayed expectation is valid as an extended nonnegative value; finiteness needs integrability of the product. The <exponential distribution> in the next part guarantees that finiteness.
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