Minimum Value.

Algebra Level 2

If x 1 , x 2 , x 3 , … , x 2019 x_1,x_2,x_3,\dots,x_{2019} are positive integers, then what is the minimum of the value of the following expression?

( x 1 + x 2 + x 3 + ⋯ + x 2019 ) ( 1 x 1 + 1 x 2 + 1 x 3 + ⋯ + 1 x 2019 ) \Big(x_1+x_2+x_3+\dots+x_{2019}\Big)\Big(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\dots+\frac{1}{x_{2019}}\Big)

2019 2019 0 0 201 9 2 2019^2 1 2019 \frac{1}{2019} 2018 2019 \frac{2018}{2019}

Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

Chew-Seong Cheong
May 13, 2019

By Hölder's inequality , we have:

( x 1 + x 2 + x 3 + ⋯ + x 2019 ) 1 2 ( 1 x 1 + 1 x 2 + 1 x 3 + ⋯ + 1 2019 ) 1 2 ≥ ( x 1 x 1 ) 1 2 + ( x 2 x 2 ) 1 2 + ( x 3 x 3 ) 1 2 + ⋯ + ( x 2019 x 2019 ) 1 2 = 2019 ⟹ ( x 1 + x 2 + x 3 + ⋯ + x 2019 ) ( 1 x 1 + 1 x 2 + 1 x 3 + ⋯ + 1 2019 ) ≥ 201 9 2 \begin{aligned} \left(x_1+x_2+x_3 + \cdots + x_{2019}\right)^\frac 12 \left(\frac 1{x_1} + \frac 1{x_2} + \frac 1{x_3} + \cdots + \frac 1{2019}\right)^\frac 12 & \ge \left(\frac {x_1}{x_1}\right)^\frac 12 + \left(\frac {x_2}{x_2}\right)^\frac 12 + \left(\frac {x_3}{x_3}\right)^\frac 12 + \cdots + \left(\frac {x_{2019}}{x_{2019}}\right)^\frac 12 = 2019 \\ \implies \left(x_1+x_2+x_3 + \cdots + x_{2019}\right) \left(\frac 1{x_1} + \frac 1{x_2} + \frac 1{x_3} + \cdots + \frac 1{2019}\right) & \ge \boxed{2019^2} \end{aligned}

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