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 $A\ge B>0$?
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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. |
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