1975 IMO Problems/Problem 1


Let $x_i, y_i$ $(i=1,2,\cdots,n)$ be real numbers such that \[x_1\ge x_2\ge\cdots\ge x_n \text{ and } y_1\ge y_2\ge\cdots\ge y_n.\] Prove that, if $z_1, z_2,\cdots, z_n$ is any permutation of $y_1, y_2, \cdots, y_n,$ then \[\sum^n_{i=1}(x_i-y_i)^2\le\sum^n_{i=1}(x_i-z_i)^2.\]


We can expand and simplify the inequality a bit, and using the fact that $z$ is a permutation of $y$, we can cancel some terms. \[\sum^n_{i=1}x_i^2 + \sum^n_{i=1}y_i^2 - 2\sum^n_{i=1}x_iy_i \leq \sum^n_{i=1}x_i^2 + \sum^n_{i=1}z_i^2 - 2\sum^n_{i=1}x_iz_i\] \[\sum^n_{i=1}x_iy_i \geq \sum^n_{i=1}x_iz_i\] Consider the pairing $x_1 \rightarrow y_1$, $x_2 \rightarrow y_2$, ... $x_n \rightarrow y_n$. By switching around some of the $y$ values, we have obtained the pairing $x_1 \rightarrow z_1$, $x_2 \rightarrow z_2$, ... $x_n \rightarrow z_n$. Suppose that we switch around two $y$-values, $y_m$ and $y_n$, such that $y_m > y_n$. If $x_m > x_n$, call this a type 1 move. Otherwise, call this a type 2 move.

Type 2 moves only increase the sum of the products of the pairs. The sum is increased by $x_n \cdot y_m + x_m \cdot y_n - x_m \cdot y_m - x_n \cdot y_n$. This is equivalent to $(x_n - x_m)(y_m - y_n)$, which is clearly nonnegative since $y_m > y_n$ and $x_n > x_m$.

We will now consider switching from the $x$-$z$ pairing back to the $x$-$y$ pairing. We will prove that from any pairing of $x$ and $z$ values, you can use just type 2 moves to navigate back to the pairing of $x$ and $y$ values, which will complete the proof.

Suppose that $x_a$ is the biggest $x$-value that is not paired with its $y$-value in the $x$ and $z$ pairing. Then, switch this $y$-value with the $y$-value currently paired with $x_a$. This is obviously a type 2 move. Continue this process until you reach back to the $x$-$y$ pairing. All moves are type 2 moves, so the proof is complete.


Solution 2

We can rewrite the summation as \[\sum^n_{i=1} x_i^2 + \sum^n_{i=1} y_i^2 - \sum^n_{i=1}x_iy_i \le \sum^n_{i=1} x_i^2 + \sum^n_{i=1} z_i^2 - \sum^n_{i=1}x_iz_i.\] Since $\sum^n_{i=1} y_i^2 = \sum^n_{i=1} z_i^2$, the above inequality is equivalent to \[\sum^n_{i=1}x_iy_i \ge \sum^n_{i=1}x_iz_i.\] We will now prove that the left-hand side of the inequality is the greatest sum reached out of all possible values of $\sum^n_{i=1}x_iz_i$. Obviously, if $x_1 = x_2 = \ldots = x_n$ or $y_1 = y_2 = \ldots = y_n$, the inequality is true. Now, assume, for contradiction, that neither of those conditions are true and that there exists some order of $z_i$s that are not ordered in the form $z_1 \ge z_2 \ge \ldots \ge z_n$ such that $\sum^n_{i=1}x_iz_i$ is at a maximum out of all possible permutations and is greater than the sum $\sum^n_{i=1}x_iy_i$. This necessarily means that in the sum $\sum^n_{i=1}x_iz_i$ there exists two terms $x_pz_m$ and $x_qz_n$ such that $x_p > x_q$ and $z_m < z_n$. Notice that \[x_pz_n + x_qz_m - (x_pz_m + x_qz_n) = (x_p-x_q)(z_n-z_m) > 0,\] which means if we make the terms $x_pz_n$ and $x_qz_m$ instead of the original $x_pz_m$ and $x_qz_n$, we can achieve a higher sum. However, this is impossible, since we assumed we had the highest sum. Thus, the inequality \[\sum^n_{i=1}x_iy_i \ge \sum^n_{i=1}x_iz_i\] is proved, which is equivalent to what we wanted to prove. \[\] ~Imajinary


It is only the most common way of rearrangement inequality after expanding and subtracting same terms.~bluesoul

See Also

1975 IMO (Problems) • Resources
Preceded by
First Question
1 2 3 4 5 6 Followed by
Problem 3
All IMO Problems and Solutions