Problem: Let $a_1$, $a_2$, \dots, $a_n$ be a sequence of $n > 3$ real numbers
such that
\[ a_1 + \dots + a_n \ge n \quad\text{and}\quad
  a_1^2 + \dots + a_n^2 \ge n^2. \]
Prove that $\max(a_1, \dots, a_n) \ge 2$.

Solution: Proceed by contradiction, assuming $a_i < 2$ for all $i$.

If all $a_i \ge 0$, then $ n^2 \le \sum_i a_i^2 < n \cdot 2^2$,
contradiction.

Otherwise, assume at least one $a_i$ is negative.
Note that if $-x$ and $-y$ are both present in the sequence ($x,y>0$),
then we can replace them with $-(x+y)$ and $0$.
So we may assume that there is exactly one negative term, say $a_n = -M$.

Now, smooth all the nonnegative $a_i$ to be $2$,
making all inequalities strict.
Now, we have that
\begin{align*}
  2(n-1) - M &> n \\
  4(n-1) + M^2 &> n^2.
\end{align*}
This gives $n-2 < M < n-2$, contradiction.

Equality in the original occurs when $n-1$ of the $a_i$ are equal to $2$
and the last one is equal to $-(n-2)$.