## Inequality with minimum function

The variables and constants are more or less based on the definitons in my previous thread: Rewrite variable in given constants and variables

There exists a stationary population $\displaystyle C$, so:
$\displaystyle \frac{C_{j+1}(t+1)}{C_j(t+1)} = \frac{C_{j+1}(t)}{C_j(t)}$, where $\displaystyle j \in \{0, ..., l-1\}$ and $\displaystyle t = \{0, 1, ...\}$
($\displaystyle C_j(t)$ means the amount of people in age group $\displaystyle j$ at time $\displaystyle t$. There are $\displaystyle l+1$ age groups, labeled from $\displaystyle 0$ to $\displaystyle l$)

Let $\displaystyle \phi = (\phi_0, \phi_1, ..., \phi_l)$ with:

$\displaystyle \phi_i = \phi_i(t) = \frac{C_i(t)}{\sum^l_{k=0} C_k(t)}$, $\displaystyle i \in \{0, 1, ..., l\}$

Now I have to show for a arbitrary population $\displaystyle A$ that is converges to the stationary distribution $\displaystyle \phi$. To prove that, let $\displaystyle A$ and $\displaystyle B$ be 2 general populations.

And let $\displaystyle m(t) = \min_{n=0,...,r} \frac{A_n(t)}{B_n(t)}$
($\displaystyle 0 < r < l$. There are $\displaystyle r + 1$ fertile age groups, labeled $\displaystyle 0$ to $\displaystyle r$)

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Show that $\displaystyle m(t) \leq m(t+1)$ by using the properties below:

$\displaystyle A_{j+1}(t+1) = A_j(t)p_j$ with $\displaystyle j \in \{0, ..., l - 1\}$
$\displaystyle A_0(t+1) = \sum^r_{k=0} A_k(t)f_k$
(these also yield for $\displaystyle B$)

and

$\displaystyle \frac{A_0(t+1)}{B_0(t+1)} = \frac{\sum^r_{k=0} A_k(t)f_k}{\sum^r_{k=0} B_k(t)f_k} = \frac{\sum^r_{k=0} \frac{A_k(t)}{B_k(t)}B_k(t)f_k}{\sum^r_{k=0} B_k(t)f_k}$

$\displaystyle p_j$ is the probability of survival for people in age group $\displaystyle j$. So the amount of people $\displaystyle A_{j+1}(t+1)$ is equal to the amount of people $\displaystyle A_j(t)$ multiplied by the probablity of survival $\displaystyle p_j$. Hence, $\displaystyle A_{j+1}(t+1) = A_j(t)p_j$

This is different for the amount of people $\displaystyle A_0$, because there are no things like $\displaystyle A_{-1}$ and $\displaystyle p_{-1}$. So this is a special case. $\displaystyle f_k$ is the fertility factor of the people in age group $\displaystyle k$. $\displaystyle k$ is between $\displaystyle 0$ and $\displaystyle r$, where $\displaystyle r < l$, because people stop 'producing' children when they get past a certain age.

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My reasoning:

$\displaystyle m(t) = \min_{n=0,...,r} \frac{A_n(t)}{B_n(t)} = \min\left\{\frac{A_0(t)}{B_0(t)}, ..., \frac{A_r(t)}{B_r(t)}\right\}$

$\displaystyle m(t+1) = \min_{n=0,...,r} \frac{A_n(t+1)}{B_n(t+1)} = \min\left\{\frac{A_0(t+1)}{B_0(t+1)}, ..., \frac{A_r(t+1)}{B_r(t+1)}\right\} =$
$\displaystyle = \min\left\{\frac{\sum^r_{k=0} A_k(t)f_k}{\sum^r_{k=0} B_k(t)f_k}, \frac{A_0(t)p_0}{B_0(t)p_0}, ..., \frac{A_{r-1}(t)p_{r-1}}{B_{r-1}(t)p_{r-1}}\right\}$

The survival probabilities of $\displaystyle m(t+1)$ are in the numerator and denominator, so they cancel each other out. So the elements in the minimumfunction of $\displaystyle m(t)$ and $\displaystyle m(t+1)$ are mostly the same.

The only two differences are that $\displaystyle m(t)$ has got the fraction $\displaystyle \frac{A_r(t)}{B_r(t)}$ and $\displaystyle m(t+1)$ hasn't.
On the other hand $\displaystyle m(t+1)$ has got $\displaystyle \frac{\sum^r_{k=0} A_k(t)f_k}{\sum^r_{k=0} B_k(t)f_k}$ and $\displaystyle m(t)$ hasn't.

In short, when $\displaystyle \frac{A_r(t)}{B_r(t)} \leq \frac{\sum^r_{k=0} A_k(t)f_k}{\sum^r_{k=0} B_k(t)f_k}$ for all $\displaystyle t \in \{0, 1, ...\}$, then $\displaystyle m(t) \leq m(t+1)$.

I thought that:
$\displaystyle \frac{\sum^r_{k=0} A_k(t)f_k}{\sum^r_{k=0} B_k(t)f_k} = \sum^r_{k=0} \frac{A_k(t)}{B_k(t)}$, but that doesn't help me either.

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Anyone who can lend me a hand?
- Fruitbowl.