Problem: Alice and Bazza are playing the \emph{inekoalaty game},
a two‑player game whose rules depend on a positive real number $\lambda$ which is known to both players.
On the $n$th turn of the game (starting with $n=1$) the following happens:
\begin{itemize}
  \ii If $n$ is odd, Alice chooses a nonnegative real number $x_n$ such that
  \[ x_1 + x_2 + \cdots + x_n \le \lambda n.  \]
  \ii If $n$ is even, Bazza chooses a nonnegative real number $x_n$ such that
  \[ x_1^2 + x_2^2 + \cdots + x_n^2 \le n.  \]
\end{itemize}
If a player cannot choose a suitable $x_n$, the game ends and the other player wins.
If the game goes on forever, neither player wins.
All chosen numbers are known to both players.

Determine all values of $\lambda$ for which Alice has a winning strategy
and all those for which Bazza has a winning strategy.

Solution: The answer is that Alice has a winning strategy for $\lambda > 1/\sqrt2$,
and Bazza has a winning strategy for $\lambda < 1/\sqrt2$.
(Neither player can guarantee winning for $\lambda = 1/\sqrt2$.)

We divide the proof into two parts.

\paragraph{Alice's strategy when $\lambda \ge 1/\sqrt2$.}
Consider the strategy where Alice always plays $x_{2i+1} = 0$ for $i=0, \dots, k-1$.

In this situation, when $n = 2k+1$ we have
\begin{align*}
  \sum_1^{2k} x_i &= 0 + x_2 + 0 + x_4 + \dots + 0 + x_{2k}  \\
  &\le k \cdot \sqrt{\frac{x_2^2 + \dots + x_{2k}^2}{k}}
  = \sqrt{2} \cdot k < \lambda \cdot (2k+1)
\end{align*}
and so the choices for $x_{2k+1}$ are
\[ x_{2k+1} \in [0, \lambda \cdot (2k+1) - \sqrt{2} k] \]
which is nonempty.
Hence Alice can't ever lose with this strategy.

But suppose further $\lambda > \frac{1}{\sqrt 2}$; we show Alice can win.
Choose $k$ large enough that
\[ \sqrt{2} \cdot k < \lambda \cdot (2k+1) - \sqrt{2k+2}. \]
Then on the $(2k+1)$st turn, Alice can (after playing $0$ on all earlier turns)
play a number greater than $\sqrt{2k+2}$ and cause Bazza to lose.

\paragraph{Bazza strategy when $\lambda \le 1/\sqrt2$.}
Consider the strategy where Bazza always plays $x_{2i+2} = \sqrt{2 - x_{2i+1}^2}$
for all $i = 0, \dots, k-1$ (i.e.\ that is, the largest possible value Bazza can play).

To analyze Bazza's choices on each of his turns, we first need to estimate $x_{2k+1}$.
We do this by writing
\begin{align*}
  \lambda \cdot (2k+1) &\ge x_1 + x_2 + \dots + x_{2k+1} \\
  &= \left(x_1 + \sqrt{2-x_1^2}\right) + \left(x_3 + \sqrt{2-x_3^2}\right) \\
  &\qquad+ \dots + \left(x_{2k-1} + \sqrt{2-x_{2k}^2}\right) + x_{2k+1} \\
  &\ge {\underbrace{\sqrt{2} + \dots + \sqrt{2}}_{k \text{ times}}} + x_{2k+1}
  = \sqrt 2 \cdot k + x_{2k+1}
\end{align*}
where we have used the fact that $t + \sqrt{2-t^2} \ge 2$ for all $t \ge 0$.
This means that
\[ x_{2k+1} \le \lambda \cdot (2k+1) - \sqrt 2 k < \sqrt 2. \]
And $x_1^2 + x_2^2 + \dots + x_{2k}^2 + x_{2k+1}^2 = (2 + \dots + 2) + x_{2k+1}^2$,
Bazza can indeed choose $x_{2k+2} = \sqrt{2-x_{2k+1}^2}$ and always has a move.

But suppose further $\lambda < 1/\sqrt2$.
Then the above calculation \emph{also} shows that Alice couldn't
have made a valid choice for large enough $k$,
since $\lambda \cdot (2k+1) - \sqrt 2 k < 0$ for large $k$.

\begin{remark*}
  In the strategies above,
  we saw that Alice prefers to always play $0$
  and Bazza prefers to always play as large as possible.
  One could consider what happens in the opposite case:
  \begin{itemize}
    \ii If Alice tries to always play the largest number possible,
    her strategy still wins for $\lambda > 1$.
    \ii If Bazza tries to always play $0$,
    Alice can win no matter the value for $\lambda > 0$.
  \end{itemize}
\end{remark*}