[Math] If $x$ and $y$ are positive numbers, what is the minimum possible value of $(x+y)\left(\frac1x + \frac1y\right)?$

Question

If $x$ and $y$ are positive numbers, what is the minimum possible value of $(x+y)\left(\frac1x + \frac1y\right)?$

Any help would be greatly appreciated. Thanks!

Answer 1

When $x=1,y=1$ then $(x+y)(\dfrac{1}{x} +\dfrac{1}{y})=4$.

To find the lower bound we will apply the basic inequality $a^2+b^2\geq 2ab$. \begin{align*} (x+y)(\dfrac{1}{x} +\dfrac{1}{y})&\geq( 2\sqrt{xy} )(2\frac{1}{\sqrt{xy}})\\ &=4 \end{align*}