Lopici A.L.

The sequence $x_{0}, x_{1}, x_{2}, \ldots$ is defined by the following conditions: $x_{0}=1, x_{1}=\lambda$, for any $n>1$ the equality holds

$$
(\alpha+\beta)^{n} x_{n}=\alpha^{n} x_{n} x_{0}+\alpha^{n-1} \beta x_{n-1} x_{1}+\alpha^{n-2} \beta^{2} x_{n-2} x_{2}+\ldots+\beta^{n} x_{0} x_{n} .
$$

Here $\alpha, \beta, \lambda$ are given positive numbers. Find $x_{n}$ and determine for which $n$ the value of $x_{n}$ is the largest.
We will prove by induction that $x_{n}=\frac{\lambda^{n}}{n!}$. The base case $(n=0,1)$ is given in the condition.

Induction step. Suppose $x_{k}=\frac{\lambda^{k}}{k!}$ for all $k \leq n-1$. Then from the equalities

$$
\left((\alpha+\beta)^{n}-\alpha^{n}-\beta^{n}\right) x_{n}=\sum_{k=1}^{n-1} \frac{\alpha^{n-k} \beta^{k} \lambda^{n}}{(n-k)!k!}=\sum_{k=1}^{n-1} \frac{\alpha^{n-k} \beta^{k} \lambda^{n}}{(n-k)!k!}=\left((\alpha+\beta)^{n}-\alpha^{n}-\beta^{n}\right) \cdot \frac{\lambda^{n}}{n!}
$$

it follows that $x_{n}=\frac{\lambda^{n}}{n!}$, since the common factor on both sides is positive for $n>1$.

The inequality $\frac{\lambda^{n-1}}{(n-1)!} \leq \frac{\lambda^{n}}{n!}$ is equivalent to the inequality $n \leq \lambda$. Thus, when transitioning from $n-1$ to $n$, the number $x_{n}$ increases if

$n \leq \lambda$. Therefore, the maximum value of $x_{n}$ is achieved at $n=[\lambda]$; if $\lambda$ is an integer, then $x_{\lambda-1}=x_{\lambda}$ are the two largest numbers in the sequence $x_{n}$.

## Answer

$x_{n}=\frac{\lambda^{n}}{n!} ;$ at $n=[\lambda]$

Prove that if

a) $a, b$ and $c$ are positive numbers, then $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$;

b) $a, b, c$ and $d$ are positive numbers, $\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}+\frac{d}{a+b+c} \geq \frac{4}{3}$;

c) $a_{1}, \ldots, a_{n}$ are positive numbers ($n>1$), then

$\frac{a_{1}}{a_{2}+a_{3}+\ldots+a_{n}}+\frac{a_{2}}{a_{1}+a_{3}+\ldots+a_{n}}+\ldots+\frac{a_{n}}{a_{1}+a_{2}+\ldots+a_{n-1}} \geq \frac{n}{n-1}$.

## Solution

c). Let $s=a_{1}+a_{2}+\ldots+a_{n}, b_{i}=s-a_{i}$. We need to prove that

$\frac{s-b_{1}}{b_{1}}+\frac{s-b_{2}}{b_{2}}+\ldots+\frac{s-b_{n}}{b_{n}} \geq \frac{n}{n-1}$, that is,

$s\left(\frac{1}{b_{1}}+\frac{1}{b_{2}}+\ldots+\frac{1}{b_{n}}\right) \geq \frac{n}{n-1}+n=\frac{n^{2}}{n-1}$.

First method. According to the inequality between the arithmetic mean and the harmonic mean (see problem $\underline{61402}$ in)

$\frac{1}{b_{1}}+\frac{1}{b_{2}}+\ldots+\frac{1}{b_{n}} \geq \frac{n^{2}}{b_{1}+b_{2}+\ldots+b_{n}}$

But $b_{1}+b_{2}+\ldots+b_{n}=n s-a_{1}-a_{2}-\ldots-a_{n}=(n-1) s$, hence

$s\left(\frac{1}{b_{1}}+\frac{1}{b_{2}}+\ldots+\frac{1}{b_{n}}\right) \geq \frac{s n^{2}}{n(n-1)}=\frac{n^{2}}{n-1}$.

Second method. $(n-1) \frac{s-b_{1}}{b_{1}}=\frac{b_{1}+b_{2}+\ldots+b_{n}-(n-1) b_{1}}{b_{1}}=\frac{b_{2}+\ldots+b_{n}}{b_{1}}-(n-2)$.

Transform the other terms $(n-1) \frac{s-b_{i}}{b_{i}}$ in the same way and sum all $n$ terms. Then group the fractions $b_{i / b_{j}}$ and $_{b_{j} / b_{i}}$ in pairs and use the inequality $b_{i /} b_{j}+b_{j / b_{i}} \geq 2$.

Since there will be $C_{n}^{2}$ such pairs, we will get:

$(n-1)\left(\frac{s-b_{1}}{b_{1}}+\ldots+\frac{s-b_{n}}{b_{n}}\right) \geq C_{n}^{2} \cdot 2-(n-2) n=n(n-1)-n^{2}+2 n=n$.