In terms of $n\ge2$, find the largest constant $c$ such that for all nonnegative $a_1,a_2,\ldots,a_n$ satisfying $a_1+a_2+\cdots+a_n=n$, the following inequality holds:
\[\frac1{n+ca_1^2}+\frac1{n+ca_2^2}+\cdots+\frac1{n+ca_n^2}\le \frac{n}{n+c}.\]

[i]Calvin Deng.[/i]
1. **Define the problem and introduce the constant \( d \):**
   We need to find the largest constant \( c \) such that for all nonnegative \( a_1, a_2, \ldots, a_n \) satisfying \( a_1 + a_2 + \cdots + a_n = n \), the inequality
   \[
   \frac{1}{n + c a_1^2} + \frac{1}{n + c a_2^2} + \cdots + \frac{1}{n + c a_n^2} \leq \frac{n}{n + c}
   \]
   holds. Let \( d = \frac{n}{c} \). We will prove, using induction, that for all \( d \geq n + \frac{1}{n-1} \), the equivalent inequality
   \[
   \sum_{i=1}^{n} \frac{1}{d + a_i^2} \leq \frac{n}{d + 1}
   \]
   holds.

2. **Base case \( n = 2 \):**
   We need to prove that for \( a + b = 2 \) and \( d \geq 3 \), the inequality
   \[
   \frac{1}{d + a^2} + \frac{1}{d + b^2} \leq \frac{2}{d + 1}
   \]
   holds. This is equivalent to:
   \[
   2a^2b^2 + d(a^2 + b^2 - 2) - a^2 - b^2 \geq 0
   \]
   Since the left-hand side (LHS) is increasing in \( d \), it is enough to prove the inequality for \( d = 3 \). This simplifies to:
   \[
   2(ab - 1)^2 \geq 0
   \]
   which is always true.

3. **Inductive step:**
   Suppose the statement holds for \( n \). We will prove it for \( n + 1 \), i.e.,
   \[
   \sum_{i=1}^{n+1} \frac{1}{d + a_i^2} \leq \frac{n+1}{d + 1}
   \]
   for \( d \geq n + 1 + \frac{1}{n} \). Without loss of generality (WLOG), assume \( a_1 \leq a_2 \leq \cdots \leq a_{n+1} \). Then \( a_1 + \cdots + a_n = tn \) for some \( t \leq 1 \).

4. **Define new variables:**
   Let \( b_i = \frac{a_i}{t} \) for all \( i = 1, \ldots, n \), so \( b_1 + \cdots + b_n = n \). Define \( d' = \frac{d}{t^2} \). Since \( d \geq n + 1 + \frac{1}{n} \), we have \( d' > n + \frac{1}{n-1} \).

5. **Apply the induction hypothesis:**
   By the induction hypothesis, it follows that:
   \[
   \sum_{i=1}^{n} \frac{1}{d' + b_i^2} \leq \frac{n}{d' + 1}
   \]
   Therefore:
   \[
   \sum_{i=1}^{n} \frac{1}{d + a_i^2} = \frac{1}{t^2} \sum_{i=1}^{n} \frac{1}{d' + b_i^2} \leq \frac{n}{d + t^2}
   \]

6. **Final step:**
   We need to prove:
   \[
   \frac{n}{d + t^2} + \frac{1}{d + a_{n+1}^2} \leq \frac{n+1}{d + 1}
   \]
   Using \( a_{n+1} = n + 1 - nt \), this is equivalent to:
   \[
   n(n+1)(t-1)^2(d + nt^2 - 2t - n - 1) \geq 0
   \]
   Since \( d \geq n + 1 + \frac{1}{n} \), the left-hand side (LHS) is:
   \[
   \text{LHS} \geq (n+1)(t-1)^2(nt-1)^2 \geq 0
   \]
   Hence, the inequality is proved. Equality holds when \( a_1 = \cdots = a_n = 1 \) and when \( a_1 = \cdots = a_{n-1} = \frac{1}{n-1}, \; a_n = n-1 \).

The final answer is \( \boxed{ c = \frac{n}{n + \frac{1}{n-1}} } \).