China-TST-2023B

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, $$\begin{gathered} \left(\lambda \sum _{i=1}^m\sqrt{c_i+b}-\sum _{i=1}^m\sqrt{c_i}\right)-\left(\lambda \sqrt{\sum _{i=1}^m{c}_i+b}-\sqrt{\sum _{i=1}^m{c}_i}\right) \\ \ge \sum _{i=1}^m(\sqrt{c_i+b}-\sqrt{c_i})-\left(\sqrt{\sum _{i=1}^m{c}_i+b}-\sqrt{\sum _{i=1}^m{c}_i}\right) \\ =\sum _{i=1}^m\frac{b}{\sqrt{c_i+b}+\sqrt{c_i}}-\frac{b}{\sqrt{{\sum }_{i=1}^m{c}_i+b}+\sqrt{{\sum }_{i=1}^m{c}_i}}\ge 0. \end{gathered}$$ 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 $$\begin{gathered} \lambda \sqrt{a-b}+\lambda (\sqrt{c+b}-\sqrt{c+2b-a})>(\sqrt{a}-\sqrt{b})+(\sqrt{c}-\sqrt{c-a+b}) \\ ⟺\lambda +\frac{\lambda \sqrt{a-b}}{\sqrt{c+b}+\sqrt{c+2b-a}}>\frac{\sqrt{a-b}}{\sqrt{a}+\sqrt{b}}+\frac{\sqrt{a-b}}{\sqrt{c}+\sqrt{c-a+b}}. \end{gathered}$$ 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 $$(2)\frac{\lambda \sqrt{a-b}}{\sqrt{c+b}+\sqrt{c+2b-a}}>\frac{\sqrt{a-b}}{\sqrt{c}+\sqrt{c-a+b}}.$$ 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 $$\lambda \sqrt{n}\ge n-1+\sqrt{n-1}⟹\lambda \ge \frac{n-1+\sqrt{n-1}}{\sqrt{n}}={\lambda }_n.$$ 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 $$(3){\lambda }_n\sum _{i=1}^n\sqrt{|a_i-b|}+\sqrt{n\left|\sum _{i=1}^n{a}_i\right|}\ge \sum _{i=1}^n\sqrt{|a_i|}.$$ Let $t:=\frac{1}{n}{\sum }_{i=1}^n{a}_i$, $a_i^{\prime }:=a_i-t$ ($i=1,2,\dots ,n$), and $b^{\prime }:=b-t$. Note that $$\sqrt{|t|}+\sqrt{|a_i^{\prime }|}\ge \sqrt{|t|+|a_i^{\prime }|}\ge \sqrt{|t+a_i^{\prime }|}=\sqrt{|a_i|}.$$ 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 $$(4){\lambda }_n\sum _{i=1}^n\sqrt{|a_i-b|}\ge \sum _{i=1}^n\sqrt{|a_i|}.$$ 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 $$(5){\lambda }_n\left(\sum _{i=1}^{n-1}\sqrt{|a_i-b|}+\sqrt{c+b}\right)\ge \sum _{i=1}^{n-1}\sqrt{a_i}+\sqrt{c}.$$ If $b\ge \frac{c}{n-1}$, then $$\begin{gathered} {\lambda }_n\left(\sum _{i=1}^{n-1}\sqrt{|a_i-b|}+\sqrt{c+b}\right)\ge {\lambda }_n\sqrt{c+\frac{c}{n-1}}=\sqrt{(n-1)c}+\sqrt{c} \\ =\sqrt{(n-1)\sum _{i=1}^{n-1}{a}_i}+\sqrt{c}\ge \sum _{i=1}^{n-1}\sqrt{a_i}+\sqrt{c}. \end{gathered}$$ 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 $\frac{n-1+\sqrt{n-1}}{\sqrt{n}}$. $\square$