Sum of fifty positive nos. is 1.
Find maximum value of sum of their inverse.
I have no idea how to solve this question... do not mark it as off topic or anything...
Maybe we should use AM>=GM?
Sum of fifty positive nos. is 1.
Find maximum value of sum of their inverse.
I have no idea how to solve this question... do not mark it as off topic or anything...
Maybe we should use AM>=GM?
Let $x_1,x_2,\ldots,x_{50}$ positive numbers such that their sum is $1$.
We can prove that the sum $\frac{1}{x_1}+\frac{1}{x_2}+\ldots+\frac{1}{x_{50}}$ is not bounded.
Let $M>1$, then by putting $x_1=\frac{1}{2M}$ and $x_2=x_3=\dots =x_{50}=\frac{1}{49}\left(1-\frac{1}{2M}\right)$ we get $$\frac{1}{x_1}+\frac{1}{x_2}+\ldots+\frac{1}{x_{50}}=2M+\underbrace{\frac{98}{2M-1}+\frac{98}{2M-1}+\ldots+\frac{98}{2M-1}}_{49\text{ times}}>M$$
The maximum is infinity. Take one of the number to approach $0$ and all the others sum to approach $1$.
Assume that the numbers are equal. Therefore the number must be $1/50$ When you say inverse if you mean reciprocal then the maximum value of the reciprocal becomes $50*50=2500$ Science I assumed the numbers are equal their sums are the maximum