Your claimed result is not true, which probably explains why you're having trouble seeing it.
For simplicity I'll let $a = 0, b = 1$. Results for general $a$ and $b$ can be obtained by a linear transformation.
Let $X_1, \ldots, X_n$ be independent uniform $(0,1)$; let $Y$ be their minimum and let $X$ be their maximum. Then the probability that $X \in [x, x+\delta x]$ and $Y \in [y, y+\delta y]$, for some small $\delta x$ and $\delta y$, is
$$ n(n-1) (\delta x) (\delta y) (x-y)^{n-2} $$
since we have to choose which of $X_1, \ldots, X_n$ is the smallest and which is the largest; then we need the minimum and maximum to fall in the correct intervals; then finally we need everything else to fall in the interval of size $x-y$ in between. The joint density is therefore $f_{X,Y}(x,y) = n(n-1) (x-y)^{n-2}$.
Then the density of $Y$ can be obtained by integrating. Alternatively, $P(Y \ge y) = (1-y)^n$ and so $f_Y(y) = n(1-y)^{n-1}$.
The conditional density you seek is then
$$ f_{X|Y}(x|y) = {n(n-1) (x-y)^{n-2} \over n(1-y)^{n-1}} == {(n-1) (x-y)^{n-2} \over (1-y)^{n-1}}. $$
where of course we restrict to $x > y$.
For a numerical example, let $n = 5, y = 2/3$. Then we get $f_{X|Y}(x/y) = 4 (x-2/3)^3 / (1/3)^4 = 324 (x-2/3)^3$ on $2/3 \le x \le 1$. This is larger near $1$ than near $2/3$, which makes sense -- it's hard to squeeze a lot of points in a small interval!
The result you quote holds only when $n = 2$ -- if I have two IID uniform(0,1) random variables, then conditional on a choice of the minimum, the maximum is uniform on the interval between the minimum and 1. This is because we don't have to worry about fitting points between the minimum and the maximum, because there are $n - 2 = 0$ of them.
Suppose the maximum is $X_{500}$, then $$P(X_{500}\le x)=P(X_i \le x ,i=1,2,...,500)$$
Note that this is so because if the maximum is less than $x$ , then every other order statistic is less than $x$.
Now since the $X_i's$ are IID, it follows that;
$$P(X_{500}\le x)=\prod_{i=1}^{500} P(X_i\le x)=x^{500}$$
which is the CDF and so the PDF is $500x^{499}$ (which is obtained by differentiation).
Now the expected value of the maximum is found as follows;
$$E[X]=\int _0^1 x (500x^{499})dx=\int _0^1 500x^{500}dx=\frac {500}{501}$$
Best Answer
You're absolutely right about being cautious regarding multiplying non-independent PDFs. However, what you can instead do is write down the joint distribution of the order statistics (in this case the first order statistic is the minimum of your three variables and the third order statistic is the max).
The joint PDF is surprisingly simple in this case $f_{\min(X_1,X_2,X_3),\max(X_1,X_2,X_3)}(u,v)=3!(v-u)$, it's essentially given by the number of possible permutations (that's the $3!$ term) times the probability that the "middle" variable is somewhere between $u$ and $v$. Check out https://en.wikipedia.org/wiki/Order_statistic#The_joint_distribution_of_the_order_statistics_of_the_uniform_distribution for more details.
From this, one approach could be to compute the expected value using multivariable calculus $$E[\min(X_1,X_2,X_3) \max(X_1,X_2,X_3)] = 3! \int_{0<u<v<1} u v (v-u) dvdu = 3! \int_0^1 \left(\int_u^1 u v (v-u) dv\right)du.$$