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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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17 views and no comment yet! Any hint? Is it too simple or too hard? – Helium Jun 8 '11 at 23:26
too many variables... where does the problem origin from? – Jun 8 '11 at 23:42
You can fix $ A \ge B > 0 $ and define the function $ f(u) = \ln (1+u/A) / \ln (1+u/B) $. If you prove $ f'(u) > 0 $ for all $ u > 0 $, you'll have demonstrated it is monotonic increasing and therefore $ X \ge Y > 0 \implies f(X) \ge f(Y) $ which entails your inequality up there. – anon Jun 8 '11 at 23:47
@user11848: It's related to the speed of computer networks :) Not an assignment, that's my research. – Helium Jun 8 '11 at 23:54
@anon: Thanks. It really works. – Helium Jun 9 '11 at 0:12
up vote 3 down vote accepted

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.


$$f_\alpha'(u) = \frac{\alpha(1+u)\log(1+u) - (1+\alpha u)\log(1+\alpha u)}{(1+u)(1+\alpha u)\log^2(1+u)}$$

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:


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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Thank you very much. I think your answer is quite similar to anon's comment above. But I think his approach is a simpler one. Anyways, your answer is perfect. – Helium Jun 9 '11 at 1:16
I was verifying your answer today and I found there is a problem with it. Actually, I think the numerator must be $\alpha (1+u)\log(1+u) - (1+\alpha u)\log(1+\alpha u)$ instead of the one you wrote (i.e. the first term has an $\alpha$ multiplier). I believe the general idea of your proof still holds but it needs a bit modification. I'll get back if I found the answer. But write here if you have any idea. – Helium Jun 20 '11 at 21:43
@Mohsen You are right, I'll see what I can do to fix that. – Thomas Andrews Jun 22 '11 at 15:01
It's perfect. Thank you. – Helium Jun 22 '11 at 19:27

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 $v_t$.

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