Question

If $x_i>0,i=1,2,....,50and{x}_1+x_2+...+x_{50}=50$ then the minimum value of $\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_{50}}$ equals to

• A. $50$
• B. ${\left(50\right)}^2$
• C. ${\left(50\right)}^3$
• D. ${\left(50\right)}^4$

Answer 1

The correct option is A $50$

Using $AM\ge HM$, we get

$\frac{x_1+x_2+x_3+.......+x_{50}}{50}\ge \frac{50}{\frac{1}{x_1}+\frac{1}{x_2}+............+\frac{1}{x_{50}}}$

$\Rightarrow \frac{50}{50}\ge \frac{50}{\frac{1}{x_1}+\frac{1}{x_2}+............+\frac{1}{x_{50}}}$, use the given values

$\therefore \frac{1}{x_1}+\frac{1}{x_2}+........+\frac{1}{x_{50}}\ge 50$