32. Given that $a b c d e$ are real numbers such that
$$
a+b+c+d+e=8
$$
and
$$
a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16
$$

Determine the maximum value of $\lfloor e\rfloor$.
32. Answer: 3
We shall apply the following inequality:
$$
4\left(a^{2}+b^{2}+c^{2}+d^{2}\right) \geq(a+b+c+d)^{2}
$$

Since $a+b+c+d=8-e$ and $a^{2}+b^{2}+c^{2}+d^{2}=16-e^{2}$, we have
$$
4\left(16-e^{2}\right) \geq(8-e)^{2}
$$
i.e. $e(5 e-16) \leq 0$. Thus $0 \leq e \leq 165$.

Note that if $a=b=c=d=65$, we have $e=165$. Hence $\lfloor e\rfloor=3$.