How can I show $\log(1+\frac{X}{A})\log(1+\frac{Y}{B})\ge \log(1+\frac{X}{B})\log(1+\frac{Y}{A})$? How can I show 
$$\log(1+\frac{X}{A})\log(1+\frac{Y}{B})\ge \log(1+\frac{X}{B})\log(1+\frac{Y}{A})$$
if $X\ge Y>0$ and $A\ge B>0$?
 A: Given $\alpha \geq 1$, define $f_\alpha(u) = \frac{\log(1+\alpha u)}{\log(1+u)}$ for $u>0.$
Show that $f_\alpha$ is an non-increasing.
Then if $\alpha = X/Y$, let $u=Y/A$ and $u'=Y/B$.  So $u\leq u'$, and, since $f_\alpha$ is non-increasing, $f_\alpha(u)\geq f_\alpha(u')$.  Expand that out, you get:
$$\frac{\log(1+X/A)}{\log(1+Y/A)} \geq \frac{\log(1+X/B)}{\log(1+Y/B)}$$
which is equivalent to the result you want.
So you only need to show that $f_\alpha$ is non-increasing, or, alternatively, that $f_\alpha'(u)\leq 0$ for all u>0.
But:
$$f_\alpha'(u) = \frac{\alpha(1+u)\log(1+u) - (1+\alpha u)\log(1+\alpha u)}{(1+u)(1+\alpha u)\log^2(1+u)}$$
So you need to show that $\alpha(1+u)\log(1+u) \leq (1+\alpha u)\log(1+\alpha u)$ when $\alpha\geq 1$ and $u > 0$.
This can be thought of as saying that $$g_u(z)=\frac{(1+uz)\log(1+uz)}{z}$$ has the property that $g_u(1)\leq g_u(\alpha)$ when $\alpha\geq 1$.
Now we show that $g_u$ is non-decreasing.
$$g_u'(z) = -\frac{(1+uz)\log(1+uz)}{z^2} + \frac{u\log(1+uz) + u}{z}$$
which simplifies to:
$$\frac{uz-\log(1+uz)}{z^2}$$
But $w\geq\log(1+w)$ for all $w\geq 0$.  Setting $w=uz$, we see that $g_u'(z)$ is non-negative when $z>0$, and thus, in particular, $g_u(1)\leq g_u(\alpha)$ when $\alpha\geq 1$ and $u>0$.  So $f_u'$ is non-positive, so $f_u$ is non-increasing, and we are done.
A: A solution of a somewhat different flavor:
Set $a=Y/A \leq b=Y/B$ and $c=X/Y\geq 1$.
Define $$v_t=\left( \begin{matrix} \log(1+ta) \\ \log(1+tb) \end{matrix} \right) \ \ \mbox{with} \ \ \ v_t' = \left( \begin{matrix} \frac{a}{1+at} \\ \frac{b}{1+bt} \end{matrix} \right) .$$
The original problem amounts to showing that $S=\det\left( v_c,v_1\right) \geq 0$.
Now, 
$$ S=\int_1^c \det(v_t',v_1) dt = \int_1^c \int_0^1 \det(v_t',v_u') du \, dt,$$
where I used that $\det(v_1,v_1)=0$ and $v_0$ vanishes.
But for $t\geq u$: 
$$\det (v_t',v_u') = \frac{ab (t-u)(b-a)}{(1+ta)(1+tb)(1+ua)(1+ub)} \geq 0.$$
The last relates to concavity of the curve traced by $v_t$.
