China National Team Selection Test

Denote $\lambda ={\left(\frac{n}{n-1}\right)}^{n-1}$, $n\ge 2$. Obviously, $\lambda >1$. For given $i\in \{1,\dots ,n\}$, fix $p=a_k^2+b_k^2$ for $k=1,2,\dots ,n$ and fix $a_j$ and $b_j$ ($j\ne i$). Then the left-hand side of $$\stackrel{◯}{1}=\frac{\lambda }{n}(p-b_i^2+\sum _{j\ne i}{a}_j^2)+\frac{1}{n^2}(b_i+\sum _{j\ne i}{b}_j{)}^2$$ is a quadratic function of $b_i$, $b_i\in [0,\sqrt{p}]$, with leading coefficient $-\frac{\lambda }{n}+\frac{1}{n^2}<0$. Thus, its minimum is taken at the endpoints, that is, $b_i=0$ or $a_i=0$. So, we can suppose that $a_i{b}_i=0,i=1,2,\dots ,n$.

Case 1. Each $a_i=0$, then by the mean value inequality, we have $${\left(\frac{1}{n}\sum _{i=1}^n{b}_i\right)}^2\ge \prod _{i=1}^n{b}_i^{\frac{2}{n}}.$$

Case 2. Each $b_i=0$, then by the mean value inequality, we have $$\lambda \left(\frac{1}{n}\sum _{i=1}^n{a}_i^2\right)\ge \frac{1}{n}\sum _{i=1}^n{a}_i^2\ge \prod _{i=1}^n{a}_i^{\frac{2}{n}}.$$

Case 3. We may suppose that $b_1=\cdots =b_k=0$, $a_{k+1}=\cdots =a_n=0$, $1\le k<n$. Let $a_1{a}_2\cdots a_k=a^k,b_{k+1}\cdots b_n=b^{n-k},a,b\ge 0$. Then by the mean value inequality, we have $$a_1^2+a_2^2+\cdots +a_k^2\ge k{a}^2,b_{k+1}+\cdots +b_n\ge (n-k)b.$$ It suffices to prove that $$\frac{\lambda k}{n}{a}^2+\frac{(n-k{)}^2}{n^2}{b}^2\ge a^{\frac{2k}{n}}\cdot b^{\frac{2(n-k)}{n}}.\stackrel{◯}{2}$$ By the mean value inequality, we see that The left-hand side of $$\begin{gathered} \stackrel{◯}{2}=\frac{\lambda }{n}{a}^2+\cdots +\frac{\lambda }{n}{a}^2+\underset{k\text{ terms}}{\underbrace{\frac{n-k}{n^2}{b}^2}}+\cdots +\underset{n-k\text{ terms}}{\underbrace{\frac{n-k}{n^2}{b}^2}} \\ \ge {\lambda }^{\frac{k}{n}}{a}^{\frac{2k}{n}}\cdot {\left(\frac{n-k}{n}\right)}^{\frac{n-k}{n}}\cdot b^{\frac{2(n-k)}{n}}. \end{gathered}$$ So, it suffices to show that ${\lambda }^{\frac{k}{n}}{\left(\frac{n-k}{n}\right)}^{\frac{n-k}{n}}\ge 1$, that is, to show ${\left(\frac{n}{n-k}\right)}^{n-k}\le {\lambda }^k$. In fact, $$\begin{array}{rl}& \underset{n-k\text{ terms}}{\underbrace{\frac{n}{n-k}\cdot \frac{n}{n-k}\cdot \cdots \cdot \frac{n}{n-k}}}\cdot \underset{n-k\text{ terms}}{\underbrace{1\cdot 1\cdot \cdots \cdot 1}}\\ & \le {\left(\frac{n+(nk-n)}{nk-k}\right)}^{nk-k}\\ & ={\left(\frac{n}{n-1}\right)}^{(n-1)k}={\lambda }^k.\square \end{array}$$