Maths Olympiad Prep

Library / /412 of 520

Algebra Difficulty 7.3 National olympiad, round 2 Prove it

Consider the inequality
∣z1′−z2′∣+∣z2′−z3′∣+⋯+∣zn−1′−zn′∣⩽λ⋅[∣z1−z2∣+∣z2−z3∣+⋯+∣zn−1−zn∣]\begin{aligned} & \left|z_{1}^{\prime}-z_{2}^{\prime}\right|+\left|z_{2}^{\prime}-z_{3}^{\prime}\right|+\cdots+\left|z_{n-1}^{\prime}-z_{n}^{\prime}\right| \\ \leqslant & \lambda \cdot\left[\left|z_{1}-z_{2}\right|+\left|z_{2}-z_{3}\right|+\cdots+\left|z_{n-1}-z_{n}\right|\right] \end{aligned}

for all complex numbers z1,z2,⋯ ,zn z_{1}, z_{2}, \cdots, z_{n} that are not all equal. Here, kzk′=∑j=1kzj,k=1,2,⋯ ,n k z_{k}^{\prime}=\sum_{j=1}^{k} z_{j}, k=1, 2, \cdots, n . Prove that: λ⩾1−1n\lambda \geqslant 1-\frac{1}{n}. When does equality hold?

Solution

7.
∑k=1n−1∣zk′−zk+1′∣=∑k=1n−1∣1k∑j=1kzj−1k+1∑j=1k+1zj∣=∑k=1n−11k(k+1)∣∑j=1kj(zj−zj+1)∣⩽∑k=1n−1(1k(k+1)⋅∑j=1kj∣zj−zj+1∣)=∑j=1n−1∑k=jn−1(1k(k+1)j∣zj−zj+1∣)=∑j=1n−1(1−jn)∣zj−zj+1∣⩽∑j=1n−1(1−1n)∣zj−zj+1∣=(1−1n)∑j=1n−1∣zj−zj+1∣\begin{aligned} \sum_{k=1}^{n-1}\left|z_{k}^{\prime}-z_{k+1}^{\prime}\right| & =\sum_{k=1}^{n-1}\left|\frac{1}{k} \sum_{j=1}^{k} z_{j}-\frac{1}{k+1} \sum_{j=1}^{k+1} z_{j}\right| \\ & =\sum_{k=1}^{n-1} \frac{1}{k(k+1)}\left|\sum_{j=1}^{k} j\left(z_{j}-z_{j+1}\right)\right| \\ & \leqslant \sum_{k=1}^{n-1}\left(\frac{1}{k(k+1)} \cdot \sum_{j=1}^{k} j\left|z_{j}-z_{j+1}\right|\right) \\ & =\sum_{j=1}^{n-1} \sum_{k=j}^{n-1}\left(\frac{1}{k(k+1)} j\left|z_{j}-z_{j+1}\right|\right) \\ & =\sum_{j=1}^{n-1}\left(1-\frac{j}{n}\right)\left|z_{j}-z_{j+1}\right| \\ & \leqslant \sum_{j=1}^{n-1}\left(1-\frac{1}{n}\right)\left|z_{j}-z_{j+1}\right| \\ & =\left(1-\frac{1}{n}\right) \sum_{j=1}^{n-1}\left|z_{j}-z_{j+1}\right| \end{aligned}

It is easy to see that equality holds when z1≠z2,z2=z3=⋯=znz_{1} \neq z_{2}, z_{2}=z_{3}=\cdots=z_{n}, hence the proposition is proved.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.