Problem: Let $n$ be an odd positive integer,
and let $x_1$, $x_2$, \dots, $x_n$ be nonnegative real numbers.
Show that
\[ \min(x_i^2+x_{i+1}^2) \le \max(2x_jx_{j+1}) \]
where $1\le i,j\le n$ and $x_{n+1}=x_1$.

Solution: It suffices to exhibit a single pair $(i,j)$ where
\[ x_i^2 + x_{i+1}^2 \le 2 x_j x_{j+1}. \]

Since $n$ is odd we can find three adjacent $x_k$'s,
call them $a$, $b$, and $c$ (with $b$ in middle) such that $a \ge b \ge c$.
Then indeed
\[ 2ab \ge b^2+c^2 \]
so we're done.

\begin{remark*}
  When $n$ is even, counterexamples to the problem statement alternate up and down,
  such as $(1,100,1,100,1,100,\dots)$.
  This provides a reason why one should consider $a \ge b \ge c$ to exploit
  the parity of $n$.
\end{remark*}