IMO 1987 SL 4

Origin: FRA

Problem

Let ABCDEFGH be a parallelepiped with AE\parallelBF\parallelCG\parallelDH. Prove the inequality AF + AH + AC \leqAB + AD + AE + AG. In what cases does equality hold?

Solution

Setting x1 = −−\to AB, x2 = −−\to AD, x3 = −\to AE, we have to prove that \parallelx1 + x2\parallel+ \parallelx2 + x3\parallel+ \parallelx3 + x1\parallel\leq\parallelx1\parallel+ \parallelx2\parallel+ \parallelx3\parallel+ \parallelx1 + x2 + x3\parallel. We have (\parallelx1\parallel+ \parallelx2\parallel+ \parallelx3\parallel)2 −\parallelx1 + x2 + x3\parallel2 = 2  1\leqi<j\leq3 (\parallelxi\parallel\parallelxj\parallel−⟨xi, xj⟩) =  1\leqi<j\leq3  (\parallelxi\parallel+ \parallelxj\parallel)2 −\parallelxi + xj\parallel2

 1\leqi<j\leq3 (\parallelxi\parallel+ \parallelxj\parallel+ \parallelxi + xj\parallel)(\parallelxi\parallel+ \parallelxj\parallel−\parallelxi + xj\parallel). The following two inequalities are obvious: \parallelxi\parallel+ \parallelxj\parallel−\parallelxi + xj\parallel\geq0, (1) \parallelxi\parallel+ \parallelxj\parallel+ \parallelxi + xj\parallel\leq\parallelx1\parallel+ \parallelx2\parallel+ \parallelx3\parallel+ \parallelx1 + x2 + x3\parallel. (2) It follows that (\parallelx1\parallel+ \parallelx2\parallel+ \parallelx3\parallel)2 −\parallelx1 + x2 + x3\parallel2 \leq

 i=1 \parallelxi\parallel+ &&&&&  i=1 xi &&&&& ⎛ ⎝2  i=1 \parallelxi\parallel−  1\leqi<j\leq3 \parallelxi + xj\parallel ⎞ ⎠, and dividing by the positive number 3 i=1 \parallelxi\parallel+ &&&3 i=1 xi &&& we obtain  i=1 \parallelxi\parallel− &&&&&  i=1 xi &&&&& \leq2  i=1 \parallelxi\parallel−  1\leqi<j\leq3 \parallelxi + xj\parallel. The inequality is proven. Let us analyze the cases of equality. If one of the vectors is null, then equality obviously holds. Suppose that xi ̸= 0, i = 1, 2, 3. For every i, j, at least one of (1) and (2) is equality. Equality in (1) holds if and only if xi and xj are collinear with the same direction, while in (2) it holds if and only if −xk and x1 + x2 + x3 are collinear with the same direction. If not all the vectors are collinear, then there are at least two distinct pairs xi, xj, i < j, for which (2) is an equality, so at least two of xi are collinear with x1 + x2 + x3, but then so is the third; hence, the sum x1 + x2 + x3 must be 0. Thus the cases of equality are (a) the

vectors are collinear with the same direction; (b) the vectors are collinear, two of them have the same direction, say xi, xj, and \parallelxk\parallel\geq\parallelxi\parallel+ \parallelxj\parallel; (c) one of the vectors is 0; (d) their sum is 0. Second solution. The following technique, although not quite elementary, is often used to effectively reduce geometric inequalities of first degree, like this one, to the one-dimensional case. Let \sigma be a fixed sphere with center O. For an arbitrary segment d in space, and any line l, we denote by \pil(d) the length of the projection of d onto l. Consider the integral of lengths of these projections on all possible directions of OP, with P moving on the sphere:

\sigma \piOP (d) d\sigma. It is clear that this value depends only on the length of d (because of symmetry); hence A \sigma \piOP d\sigma = c \cdot |d| for some constant c ̸= 0. (1) Notice that by the one-dimensional case, for any point P \in\sigma, \piOP (x1) + \piOP (x2) + \piOP (x3) + \piOP (x1 + x2 + x3) \geq\piOP (x1 + x2) + \piOP (x1 + x3) + \piOP (x2 + x3). By integration on \sigma, using (1), we obtain c(\parallelx1\parallel+\parallelx2\parallel+\parallelx3\parallel+\parallelx1+x2+x3\parallel) \geqc(\parallelx1+x2\parallel+\parallelx1+x3\parallel+\parallelx2+x3\parallel).