Numerical solution..

Algebra Level 3

Let a 1 , a 2 , a 3 , ⋯ , a 101 a_1, a_2, a_3, \cdots, a_{101} be positive real numbers with a sum s s . Find the minimum possible value of the expression

s s − a 1 + s s − a 2 + s s − a 3 + ⋯ + s s − a 101 . \frac {s}{s-a_1} + \frac {s}{s-a_2} + \frac s{s-a_3} + \cdots +\frac {s}{s-a_{101}}.


The answer is 102.01.

Muhammad Alif by Muhammad Alif , 20, Indonesia

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1 solution

Chew-Seong Cheong
Aug 12, 2018

Using Hölder's inequality as follows:

( s s − a 1 + s s − a 2 + s s − a 3 + ⋯ + s s − a 101 ) 1 2 ( s − a 1 + s − a 2 + s − a 3 + ⋯ + s − a 101 ) 1 2 ≥ s 1 2 + s 1 2 + s 1 2 + ⋯ + s 1 2 ⏟ Number of s 1 2 = 101 Squaring both sides ( s s − a 1 + s s − a 2 + s s − a 3 + ⋯ + s s − a 101 ) ( 100 s ) ≥ 10 1 2 s s s − a 1 + s s − a 2 + s s − a 3 + ⋯ + s s − a 101 ≥ 10201 101 = 102.01 \begin{aligned} \left(\frac s{s-a_1} + \frac s{s-a_2} + \frac s{s-a_3} + \cdots + \frac s{s-a_{101}} \right)^\frac 12 \left(s-a_1 + s - a_2 + s - a_3 + \cdots + s - a_{101}\right)^\frac 12 & \ge \underbrace{s^\frac 12 + s^\frac 12 + s^\frac 12 + \cdots + s^\frac 12}_{\text{Number of }s^\frac 12 = 101} & \small \color{#3D99F6} \text{Squaring both sides} \\ \left(\frac s{s-a_1} + \frac s{s-a_2} + \frac s{s-a_3} + \cdots + \frac s{s-a_{101}} \right) \left(100s \right) & \ge 101^2s \\ \frac s{s-a_1} + \frac s{s-a_2} + \frac s{s-a_3} + \cdots + \frac s{s-a_{101}} & \ge \frac {10201}{101} = \boxed{102.01} \end{aligned}

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Really amazing solution!.

A Former Brilliant Member - 2 years, 6 months ago

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