讨论为主， 把分金沙先答案附上。

三人分金沙， 如何公平 答案有些长， 需要有耐心，稍微明白人应该可以读懂, 值得一读. 不一定人人拍案惊奇,但其 严密性是山寨解无法比拟的。

原文自wiki, 随便很多网站也有解释. 自1961年此解后,有若干书和 论文优化探讨证明(envy-free division) . 三人解法现有10多种了, 四人只能有 moving knife 的解 法. 若干教授专业于此领域.

英文不好,打字又慢, 大概翻译一下主要步骤. 请以原文为主. 找茬的乱棒打出, 自己去独英文部分好了.
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把三个人叫p1 p2 p3 . 准备好… 开始…!
(1步: P1 切蛋糕为三份)
1- P1 divides the cake into three pieces he considers of equal size. Let's call A the largest piece according to P2.
(2 步: P2 将三份中　p2 认为大的一份（A） 切下一点, 使得被切后的　A和第二大的那块相同（当然这是对于P2认为的来说的). 切下的叫A2, 悲切后的A叫　A1.
2- P2 cuts off a bit of A to make it the same size as the second largest. Now A is divided into: the trimmed piece A1 and the trimmings A2. Leave the trimmings A2 to one side.
(3, 4, 5, 6 步, 简要的说 -调整后的三大块由p3, p2, P1 的顺序来选 )
3- If P2 thinks that the two largest parts are equal, then each player chooses a part in this order: P3, P2 and finally P1.
4- P3 chooses a piece among A1 and the two other pieces.
5- P2 chooses a piece with the limitation that if P3 didn't choose A1, P2 must choose it.
6- P1 chooses the last piece leaving just the trimmings A2 to be divided.
(-恭喜!， 到此时，除了A2那小块 ，蛋糕主体已被公平分配好了(envy free). --现在来看看如何分配那一小块A2….----)
- Now, the cake minus the trimmings A2 has been divided in an envy free manner.
( 注意当初得到A1那块蛋糕的（即被p2切 后的那块）只有可能被p3, p2 其中的一个选走， 我们在此重新定义得到这个A1 的人是 PA , 另一人是PB。(P1当然还是那个老p1.)
The trimmed piece A1 has been chosen by either P2 or P3. Let's call the player who chose it PA and the other one Player PB.
(6, 7, 8步: 简要的说 -PB再切那一小块　A2为三份，然后PA, P1, PB 依次选定。 结束!)
6- PB cuts A2 into three equal pieces.
7- PA chooses a piece of A2. P1 chooses a piece of A2.
8- PB chooses the last remaining piece of A2.


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(最后关于证明不难，就留给大家娱乐了，
P1会认为他得到的〉= PA 的， 也〉= PB的;
PA会认为他得到的〉= P1 的， 也〉= PB的;
PB会认为他得到的〉= P1 的， 也〉= PA的;
let's see why the procedure is envy-free. It must be shown that each player believes that no other player received more than him. In the following analysis "largest" means "largest according to the player". Without loss of generality, we call B the piece that PB received and C the piece that P1 received.
PA received A1 plus the largest piece of A2. For him, A1 ≥ B and A1 ≥ C. So obviously, no other player received more than him.
PB received B plus a piece of A2. For him, B ≥ A1 and B ≥ C. Also, he cut A2 in 3 piece, so for him all the pieces are equal.
P1 received C plus a piece of A2. For him, C ≥ A1 and C = B.
P1 believes that PB didn't receive more than him because P1 chose his piece of A2 before PB.
P1 believes that PA didn't receive more than him because for P1, C is equal to A and A = A1 + A2; therefore C is larger than A1 plus the piece of A2 that PA received. (Even if PA took the whole A2, P1 would not envy PA.)