Problem 1100

The fourteenth question: Given a positive integer $n \geq 3$, find the largest real number $M$ such that $\sum_{k=1}^{n}\left(\frac{a_{k}}{a_{k}+a_{k+1}}\right)^{2} \geq M$ holds for any positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$, where $a_{n+1}=a_{1}$.
Solution: Let $x_{k}=\frac{a_{k+1}}{a_{k}}(1 \leq k \leq n)$, then $\prod_{k=1}^{n} x_{k}=1$. At this point, $\sum_{k=1}^{n}\left(\frac{a_{k}}{a_{k}+a_{k+1}}\right)^{2}=\sum_{k=1}^{n} \frac{1}{\left(1+x_{k}\right)^{2}}$.
(1) When $n \geq 4$, let $x_{1}=x_{2}=\ldots=x_{n-1}=\varepsilon, x_{n}=\frac{1}{\varepsilon^{n-1}}$, and let $\varepsilon \rightarrow+\infty$, then $\sum_{k=1}^{n}\left(\frac{a_{k}}{a_{k}+a_{k+1}}\right)^{2}=\sum_{k=1}^{n} \frac{1}{\left(1+x_{k}\right)^{2}} \rightarrow 1$. Hence, $M \leq 1$.
We now prove: When $n \geq 4$, $\sum_{k=1}^{n} \frac{1}{\left(1+x_{k}\right)^{2}} \geq 1$.
First, we prove a lemma: If $a, b \in \mathbb{R}^{+}$, then $\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}} \geq \frac{1}{1+a b}$.
Proof of the lemma: According to the Cauchy-Schwarz inequality, we have: $(1+a b)\left(1+\frac{a}{b}\right) \geq(1+a)^{2} \Rightarrow \frac{1}{(1+a)^{2}} \geq \frac{b}{(a+b)(1+a b)}$. Similarly, $\frac{1}{(1+b)^{2}} \geq \frac{a}{(a+b)(1+a b)}$. Therefore, $\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}} \geq \frac{b}{(a+b)(1+a b)}+\frac{a}{(a+b)(1+a b)}=\frac{1}{1+a b}$. The lemma is proved!
Now, let's return to the original problem.
When $n=4$, according to the lemma, we have:
$$\frac{1}{\left(1+x_{1}\right)^{2}}+\frac{1}{\left(1+x_{2}\right)^{2}}+\frac{1}{\left(1+x_{3}\right)^{2}}+\frac{1}{\left(1+x_{4}\right)^{2}} \geq \frac{1}{1+x_{1} x_{2}}+\frac{1}{1+x_{3} x_{4}}=\frac{1}{1+x_{1} x_{2}}+\frac{x_{1} x_{2}}{x_{1} x_{2}+x_{1} x_{2} x_{3} x_{4}}=1$$
The proposition holds.
Assume the conclusion holds for $n$. When $n+1$, without loss of generality, let $x_{n+1}=\max _{1 \leq i \leq n+1} x_{i}$, then $x_{n+1} \geq 1$. According to the induction hypothesis, we have:
$$\sum_{k=1}^{n+1} \frac{1}{\left(1+x_{k}\right)^{2}} \geq \sum_{k=1}^{n} \frac{1}{\left(1+x_{k}\right)^{2}} \geq \sum_{k=1}^{n-1} \frac{1}{\left(1+x_{k}\right)^{2}}+\frac{1}{\left(1+x_{n} x_{n+1}\right)^{2}} \geq 1$$
Therefore, when $n \geq 4$, the maximum value of $M$ is 1.
(2) When $n=3$, $x_{1}=x_{2}=x_{3}=1$, then $\sum_{k=1}^{3} \frac{1}{\left(1+x_{k}\right)^{2}}=\frac{3}{4}$, hence $M \leq \frac{3}{4}$.
We now prove: $\sum_{k=1}^{3} \frac{1}{\left(1+x_{k}\right)^{2}} \geq \frac{3}{4}$.
In fact, according to the lemma in (1), it is easy to see:
$$\frac{1}{\left(1+x_{1}\right)^{2}}+\frac{1}{\left(1+x_{2}\right)^{2}}+\frac{1}{\left(1+x_{3}\right)^{2}}+\frac{1}{(1+1)^{2}} \geq \frac{1}{1+x_{1} x_{2}}+\frac{1}{1+x_{3} \times 1}=\frac{1}{1+x_{1} x_{2}}+\frac{x_{1} x_{2}}{x_{1} x_{2}+x_{1} x_{2} x_{3}}=1$$
$\Rightarrow \sum_{k=1}^{3} \frac{1}{\left(1+x_{k}\right)^{2}} \geq \frac{3}{4}$
In summary, the maximum value of $M$ is $\left\{\begin{array}{ll}\frac{3}{4}, & \text { when } n=3 \\ 1, & \text { when } n \geq 4\end{array}\right.$.
Note 1: Given $a, b, c, d \in \mathbb{R}^{+}, a b c d=1$, prove: $\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geq 1$. (2006 China National Training Team Problem)
Proof: By completing the square, we have:
$$\begin{array}{l} \frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}-\frac{1}{1+a b}=\frac{a^{3} b+a b^{3}-a^{2} b^{2}-2 a b+1}{(1+a)^{2} \cdot(1+b)^{2} \cdot(1+a b)}=\frac{a b(a-b)^{2}+(a b-1)^{2}}{(1+a)^{2} \cdot(1+b)^{2} \cdot(1+a b)} \geq 0 \\ \Rightarrow \frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}} \geq \frac{1}{1+a b} \end{array}$$
Similarly, $\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geq \frac{1}{1+c d}$. Therefore,
$$\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geq \frac{1}{1+a b}+\frac{1}{1+c d}=\frac{c d}{c d+a b c d}+\frac{1}{1+c d}=\frac{c d}{c d+1}+\frac{1}{1+c d}=1$$
The inequality is proved.
Note 2: In fact, $(1+a b)\left(1+\frac{a}{b}\right) \geq(1+a)^{2} \Rightarrow \frac{1}{(1+a)^{2}} \geq \frac{b}{(1+a b)(a+b)}$. Similarly, $\frac{1}{(1+b)^{2}} \geq \frac{a}{(1+a b)(a+b)}$, hence $\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}} \geq \frac{1}{1+a b}$.