Since $a \le b \le c \le d \le e$, $(a-b)(b-e)+(a-c)(c-e)+(a-d)(d-e) \ge 0$. This is equivalent to $$-(b^2+c^2+d^2)+(a+e)(b+c+d)-3ae \ge 0$$$$a^2+e^2-14+3-3ae \ge 0$$$$(a+e)^2-11 \ge 5ae$$$$ae \le -2$$Equality holds when $a=b=-1$ and $c=d=e=2$, so the maximum value of $ae$ is $\boxed{-2}$.

Let $a, b, c, d, e$ be real numbers satisfying $ a \le b \le c \le d \le e, \quad a+e=1, \quad b+c+d=3, \quad a^2+b^2+c^2+d^2+e^2=14.$ Prove that$$-5\leq ae\leq -2$$

Let $a, b, c, d$ be real numbers satisfying $ a \le b \le c \le d , \quad a+d=1, \quad b+c=2, \quad a^2+b^2+c^2+d^2=9.$ Prove that$$-3\leq ad\leq -2$$