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Algebra Level 3

If x i > 0 x_i > 0 , for i = 1 , 2 , … , 50 i = 1 , 2 , \ldots , 50 and x 1 + x 2 + x 3 + … + x 50 = 50 x_1 + x_2 + x_3 + \ldots + x_{50} = 50 . Find the minimum value of 1 x 1 + 1 x 2 + … + 1 x 50 \dfrac{1}{x_1} + \dfrac{1}{x_2} + \ldots + \dfrac{1}{x_{50}} .

Inspired by lots of problem on brilliant


The answer is 50.

Akhil Bansal by Akhil Bansal , 22, India

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1 solution

Siddharth Singh
Oct 30, 2015

We can use the T2's lemma.

1 x 1 + 1 x 2 + . . . + 1 x 50 ≥ ( 50 ) 2 x 1 + x 2 + . . . + x 50 \frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{50}}\ge \frac{(50)^{2}}{x_{1}+x_{2}+...+x_{50}}

1 x 1 + 1 x 2 + . . . + 1 x 50 ≥ ( 50 ) 2 50 = 50 \frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{50}}\ge \frac{(50)^{2}}{50}=50

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oh great!! but I created this problem without thinking of titu's lemma theorem.

Akhil Bansal - 5 years, 7 months ago

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What is your solution? I solved using Cauchy-Schwarz

Kushagra Sahni - 5 years, 7 months ago

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How did you solve using cauchy schwarz ?

Sriram Venkatesan - 2 years, 7 months ago

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