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
$$\textcircled{1} = \frac{\lambda}{n} (p - b_i^2 + \sum_{j \ne i} a_j^2) + \frac{1}{n^2} (b_i + \sum_{j \ne i} b_j)^2$$
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
$$\left(\frac{1}{n} \sum_{i=1}^{n} b_i\right)^2 \geq \prod_{i=1}^{n} b_i^{\frac{2}{n}}.$$

Case 2. Each $b_i = 0$, then by the mean value inequality, we have
$$\lambda \left( \frac{1}{n} \sum_{i=1}^{n} a_i^2 \right) \geq \frac{1}{n} \sum_{i=1}^{n} a_i^2 \geq \prod_{i=1}^{n} a_i^{\frac{2}{n}}.$$

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
$$a_1^2 + a_2^2 + \cdots + a_k^2 \ge ka^2, \quad b_{k+1} + \cdots + b_n \ge (n-k)b.$$
It suffices to prove that
$$\frac{\lambda k}{n} a^2 + \frac{(n-k)^2}{n^2} b^2 \geq a^{\frac{2k}{n}} \cdot b^{\frac{2(n-k)}{n}} . \qquad \textcircled{2}$$
By the mean value inequality, we see that
The left-hand side of
$$\textcircled{2} = \frac{\lambda}{n} a^2 + \cdots + \frac{\lambda}{n} a^2 + \underbrace{\frac{n-k}{n^2} b^2}_{k \text{ terms}} + \cdots + \underbrace{\frac{n-k}{n^2} b^2}_{n-k \text{ terms}} \\ \geq \lambda^{\frac{k}{n}} a^{\frac{2k}{n}} \cdot \left( \frac{n-k}{n} \right)^{\frac{n-k}{n}} \cdot b^{\frac{2(n-k)}{n}}.$$
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,
$$\begin{align*} & \underbrace{\frac{n}{n-k} \cdot \frac{n}{n-k} \cdot \dots \cdot \frac{n}{n-k}}_{n-k \text{ terms}} \cdot \underbrace{1 \cdot 1 \cdot \dots \cdot 1}_{n-k \text{ terms}} \\ & \le \left( \frac{n + (nk - n)}{nk - k} \right)^{nk-k} \\ & = \left( \frac{n}{n-1} \right)^{(n-1)k} = \lambda^k. \quad \square \end{align*}$$