MathLabs

Problem 1

Real numbers a1,a2,…,ana_1, a_2, \ldots, a_n are given. For each ii (1≤i≤n1 \le i \le n) define di=max⁡{aj:1≤j≤i}−min⁡{aj:i≤j≤n}d_i = \max\{a_j : 1 \le j \le i\} - \min\{a_j : i \le j \le n\} and let d=max⁡{di:1≤i≤n}d = \max\{d_i : 1 \le i \le n\}. (a) Prove that, for any real numbers x1≤x2≤⋯≤xnx_1 \le x_2 \le \cdots \le x_n, max⁡{∣xi−ai∣:1≤i≤n}≥d2\max\{|x_i - a_i| : 1 \le i \le n\} \ge \dfrac{d}{2}. (b) Show that there exist real numbers x1≤x2≤⋯≤xnx_1 \le x_2 \le \cdots \le x_n such that equality holds in the inequality above.
Step 3 of 5: Conclude the lower bound of part (a)
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In plain words

If two numbers add up to at least dd, the larger of the two is at least d2\dfrac{d}{2}.

max⁡{∣xi−ai∣:1≤i≤n}≥max⁡{ap−xp, xr−ar}≥d2\max\{|x_i - a_i| : 1 \le i \le n\} \ge \max\{a_p - x_p,\, x_r - a_r\} \ge \frac{d}{2}
Detailed analysis

From (ap−xp)+(xr−ar)≥d(a_p - x_p) + (x_r - a_r) \ge d, at least one of ap−xpa_p - x_p or xr−arx_r - a_r is at least d2\dfrac{d}{2}. Therefore max⁡{∣xi−ai∣:1≤i≤n}≥max⁡{∣xp−ap∣,∣xr−ar∣}≥max⁡{ap−xp,xr−ar}≥d2\max\{|x_i - a_i| : 1 \le i \le n\} \ge \max\{|x_p - a_p|, |x_r - a_r|\} \ge \max\{a_p - x_p, x_r - a_r\} \ge \dfrac{d}{2}, proving part (a).