Problem 1935

We will prove that $\lambda_n := \frac{n-1+\sqrt{n-1}}{\sqrt{n}}$ is the desired minimum value. Let's first prove a few lemmas.

Lemma 1: The sequence $\{\lambda_n\}$ is strictly increasing.

Proof: Note that
$$\lambda_{n+1}^2 - \lambda_n^2 = \frac{n+1+2n^{\frac{5}{2}}-2(n+1)(n-1)^{\frac{3}{2}}}{n(n+1)} = \frac{n+1+2n^2\sqrt{n}-2(n^2-1)\sqrt{n-1}}{n(n+1)} > 0.$$
Therefore, $\lambda_n < \lambda_{n+1}$.

Lemma 2: Let $\lambda \ge 1$, $b > 0$, and $c_1, c_2, \dots, c_m$ be non-negative real numbers. Then
$$\lambda \sum_{i=1}^{m} \sqrt{c_i + b} - \sum_{i=1}^{m} \sqrt{c_i} \ge \lambda \sqrt{\sum_{i=1}^{m} c_i + b} - \sqrt{\sum_{i=1}^{m} c_i}.$$
Proof: It is clear that $\sum_{i=1}^{m} \sqrt{c_i + b} \ge \sqrt{\sum_{i=1}^{m} c_i + b}$. Therefore,
$$\left( \lambda \sum_{i=1}^{m} \sqrt{c_i + b} - \sum_{i=1}^{m} \sqrt{c_i} \right) - \left( \lambda \sqrt{\sum_{i=1}^{m} c_i + b} - \sqrt{\sum_{i=1}^{m} c_i} \right) \\ \ge \sum_{i=1}^{m} (\sqrt{c_i + b} - \sqrt{c_i}) - \left( \sqrt{\sum_{i=1}^{m} c_i + b} - \sqrt{\sum_{i=1}^{m} c_i} \right) \\ = \sum_{i=1}^{m} \frac{b}{\sqrt{c_i + b} + \sqrt{c_i}} - \frac{b}{\sqrt{\sum_{i=1}^{m} c_i + b} + \sqrt{\sum_{i=1}^{m} c_i}} \ge 0.$$
Hence, Lemma 2 holds.

Lemma 3: Let $\lambda \ge \lambda_2$, $b < a \le c$ be positive real numbers. Then
$$\lambda(\sqrt{a-b} + \sqrt{c+b}) - \sqrt{a} - \sqrt{c} > \lambda(\sqrt{c+2b-a} - \sqrt{b} - \sqrt{c-a+b}).$$
Proof: The inequality in Lemma 3 is equivalent to
$$\lambda\sqrt{a-b} + \lambda(\sqrt{c+b} - \sqrt{c+2b-a}) > (\sqrt{a}-\sqrt{b}) + (\sqrt{c}-\sqrt{c-a+b}) \\ \iff \lambda + \frac{\lambda\sqrt{a-b}}{\sqrt{c+b} + \sqrt{c+2b-a}} > \frac{\sqrt{a-b}}{\sqrt{a}+\sqrt{b}} + \frac{\sqrt{a-b}}{\sqrt{c}+\sqrt{c-a+b}}.$$
Firstly, we have
(1) $$\lambda > 1 > \frac{\sqrt{a-b}}{\sqrt{a} + \sqrt{b}}.$$

Furthermore, since
$$\sqrt{c+b} + \sqrt{c+2b-a} < \sqrt{c+c} + \sqrt{2(c-a)+2b} = \sqrt{2} \cdot (\sqrt{c} + \sqrt{c-a+b}),$$
and $\lambda \ge \lambda_2 = \sqrt{2}$, we have
$$(2) \qquad \frac{\lambda\sqrt{a-b}}{\sqrt{c+b} + \sqrt{c+2b-a}} > \frac{\sqrt{a-b}}{\sqrt{c} + \sqrt{c-a+b}}.$$
Therefore, combining (1) and (2), we obtain the desired inequality.

Proof: Let's return to the original problem. First, we take $a_1 = \cdots = a_{n-1} = b = 1$ and $a_n = -(n-1)$. Then, we have
$$\lambda\sqrt{n} \ge n - 1 + \sqrt{n-1} \implies \lambda \ge \frac{n-1 + \sqrt{n-1}}{\sqrt{n}} = \lambda_n.$$
Next, we will prove that $\lambda_n$ satisfies the desired inequality. That is, we need to prove that for any real numbers $a_1, a_2, \dots, a_n, b$, we have
$$(3) \qquad \lambda_n \sum_{i=1}^{n} \sqrt{|a_i - b|} + \sqrt{n \left| \sum_{i=1}^{n} a_i \right|} \ge \sum_{i=1}^{n} \sqrt{|a_i|}.$$
Let $t := \frac{1}{n} \sum_{i=1}^{n} a_i$, $a_i' := a_i - t$ ($i = 1, 2, \dots, n$), and $b' := b - t$. Note that
$$\sqrt{|t|} + \sqrt{|a_i'|} \ge \sqrt{|t| + |a_i'|} \ge \sqrt{|t + a_i'|} = \sqrt{|a_i|}.$$
Therefore, to prove (3), it suffices to prove that for any real numbers $a_1, a_2, \dots, a_n, b$, if $\sum_{i=1}^{n} a_i = 0$, then
$$(4) \qquad \lambda_n \sum_{i=1}^{n} \sqrt{|a_i - b|} \ge \sum_{i=1}^{n} \sqrt{|a_i|}.$$
Without loss of generality, let's assume that $b > 0$. By Lemma 1 and Lemma 2, we can combine all negative terms in $a_1, a_2, \dots, a_n$. Therefore, to prove (4), it suffices to prove that for any non-negative real numbers $a_1, a_2, \dots, a_{n-1}, b$, if $c := \sum_{i=1}^{n-1} a_i > 0$, then
$$(5) \qquad \lambda_n \left( \sum_{i=1}^{n-1} \sqrt{|a_i - b|} + \sqrt{c+b} \right) \ge \sum_{i=1}^{n-1} \sqrt{a_i} + \sqrt{c}.$$
If $b \ge \frac{c}{n-1}$, then
$$\lambda_n \left( \sum_{i=1}^{n-1} \sqrt{|a_i - b|} + \sqrt{c+b} \right) \ge \lambda_n \sqrt{c + \frac{c}{n-1}} = \sqrt{(n-1)c} + \sqrt{c} \\ = \sqrt{(n-1) \sum_{i=1}^{n-1} a_i} + \sqrt{c} \ge \sum_{i=1}^{n-1} \sqrt{a_i} + \sqrt{c}.$$
Therefore, (5) holds.

If $b < \frac{c}{n-1} = \frac{1}{n-1} \sum_{i=1}^{n-1} a_i$, then there must exist some $a_j > b$. By Lemma 3, after adjusting $a_j$ to $b$ and $c$ to $c - a_j + b$, the difference between the left-hand side and the right-hand side of (5) strictly decreases. After a finite number of adjustments, we will eventually reach the case where $b \ge \frac{c}{n-1}$. Therefore, (5) holds.

Combining the above, we conclude that the minimum value of $\lambda$ satisfying the given conditions is $\frac{n-1+\sqrt{n-1}}{\sqrt{n}}$. $\square$