If $x_i>0$ i=1,2,...50 and $x_1+x_2+x_3+ \cdots+x_{50}=50$, find the minimum value of $ \frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_{50}}$

Question

If $x_i>0$, for i=1,2,...50 and $x_1+x_2+x_3+ \cdots+x_{50}=50$, find the minimum value of $ \frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_{50}}$

I have no idea as to how to go about solving this. I have just tried hit and try and have realised the minimum value comes when all terms have a value of 1. I have figured out why that happens in a very basic sense:
Let $x_i$ be the only term having a value greater than one then there should be many terms with a value less than one to make the sum 50. Those "many terms" when reciprocated become greater than one while there is only one number less than one(i.e. $1/x_i$). Hence the sum of $ \frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_{50}}$ becomes larger than 50. In case all terms are taken as 1, the sum of $ \frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_{50}}$ comes out to be 50. Similarly, this can be proved for any no. of terms greater than one.

I am pretty sure this explanation is not enough to justify the answer of a maths problem but I don't know how to give a mathematical proof to this...It would be great if someone could help me solve this.

Answer 1

Hint: due to inequality of means:

$$ \frac{x_1 + \cdots + x_n}{n} \geq \frac{n}{\frac{1}{x_1} + \cdots + \frac{1}{x_n}} $$

Values are equal if and only if $x_1=x_2=\cdots=x_n$

Answer 2

As a basic example to proof the sequence take $x_{1}\dots x_{49} = 1.001$ and $x_{50} = 0.951$. The sum of all terms gives 50.

In the 2nd sequence sum up all the reciprocals:

$\frac{1}{x_{1}} + \dots + \frac{1}{x_{49}} = \frac{1}{1.001} + \dots + \frac{1}{1.001} = \frac{49}{1.001} = 48.951048951$

Now sum the last term to complete the sequence:

$\frac{49}{1.001} + \frac{1}{0.951} = 48.951048951 + 1.051524711 = 50.002573662$

As you see we end up with a value close to 50, but increasing $x_{i}$ will decrease others $x_{i'}$, so if every term is equal you will obtain the minimum as stated in the previous answer.