Let $a=(a_n)_{n\geq 1}$ and $b=(b_n)_{n\geq 1}$ be two sequences of strictly positive reals, and $x=(x_n)_{n\geq 1}$ be :
$$\forall n\in\mathbb{N}^*,\quad x_n = a_1^{b_1}a_2^{b_2}\dots a_n^{b_n}\times \left(\frac{b_1+b_2+\dots + b_n}{a_1b_1 + a_2b_2 + \dots + a_nb_n}\right)^{b_1+b_2+\dots+b_n}$$
Show that $(x_n)$ converges, and that $\forall \alpha\in [0,1]$, we can choose $a$ and $b$ such that $x_n \to \alpha$.
At the moment, all I could do was noticing that the equation is homogeneous in $a_k$ (which means replacing $a_k$ by $\lambda \cdot a_k$, $\lambda > 0$ does not change the equation).
It let us assume another condition on $a$. For example, we may assume that $a_1 + a_2 + \dots + a_n = 1$ (because by taking $a_i' = \lambda a_i$, we'll be back at $a_1 + a_2 + \dots + a_n = \lambda$, whatever $\lambda>0$ is).
Anyone would know how to solve this problem?
Edit: I managed to show that $x_n\leq 1$. To do so, we take both side to the power of $b_1+\dots+b_n$, and then we pass the $\left(\frac{b_1+b_2+\dots + b_n}{a_1b_1 + a_2b_2 + \dots + a_nb_n}\right)$ term on the left-hand side. We then recognize the Generalized Inequality of arithmetic and geometric means.
Best Answer
The key idea I see is as follows.
There are many ways to prove it. We may consider $x_{n+1}$ as a function of $a_{n+1}$ alone (that is, assuming the values of $a_1,\ldots,a_n,b_1,\ldots,b_{n+1}$ fixed, and allowing $a_{n+1}$ to vary). Clearly $x_{n+1}\to 0$ when $a_{n+1}\to 0$ or when $a_{n+1}\to\infty$. Hence there exists a value of $a_{n+1}$ at which $x_{n+1}$ attains its maximum. At any such value of $a_{n+1}$, we must have $$0=\frac{1}{x_{n+1}}\frac{\partial x_{n+1}}{\partial a_{n+1}}=\frac{b_{n+1}}{a_{n+1}}-\frac{b_{n+1}\sum_{k=1}^{n+1}b_k}{\sum_{k=1}^{n+1}a_k b_k}.$$ Solving this for $a_{n+1}$, we obtain \eqref{recurrence} (as the only solution). And indeed $x_{n+1}=x_n$ at this value of $a_{n+1}$.
So, the convergence of $x_n$ is clear. Given $\alpha\in[0,1]$, let $b_n=1$ for all $n$. Now, if $\alpha=0$, we just take $a_n$ be tending to $0$ rapidly enough as $n\to\infty$ (to have $x_n\to 0$). Otherwise, we may take $a_1=1$, solve $4a_2/(1+a_2)^2=\alpha$ for $a_2$, and determine $a_{n+1}$ for $n>1$ according to \eqref{recurrence}.