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*}

Answer 2

given real $r > 0,$ we have $$ 0 \leq \left( \sqrt r - \frac{1}{\sqrt r} \right)^2 = r -2 + \frac{1}{r}, $$ so $$ 2 \leq r + \frac{1}{r} $$ For your product, let $$ r = \frac{x}{y} $$ after you multiply it out

Answer 3

$$(x+y)\left(\frac1x+\frac1y\right)=2+\frac xy+\frac yx $$

Since $$\frac xy+\frac yx \ge 2$$ (Use AM-GM)

$$(x+y)\left(\frac1x+\frac1y\right) \ge 4$$

Answer 4

$(x+y)(\dfrac{{1}}{x}+\dfrac{{1}}{y})=1+\dfrac{{x}}{y}+1+\dfrac{{y}}{x} \geq 2 + 2 \sqrt{\frac{x}{y}\cdot\frac{y}{x}} = 4$.

Answer 5

Yet another variation on AM-GM:

$$\require{cancel} (x+y)\left(\frac1x + \frac1y\right) = \frac{(x+y)^2}{xy} \ge \frac{\left(2 \cancel{\sqrt{xy}}\right)^2}{\cancel{xy}} = 4 $$

Equality is attained when the terms are equal i.e. $\,x=y\,$.

Answer 6

By C.S.

$$(x+y)\left(\frac{1}{x}+\frac{1}{y}\right)\ge\left[(\sqrt{x})\left(\frac{1}{\sqrt{x}}\right)+(\sqrt{y})\left(\frac{1}{\sqrt{y}}\right)\right]^2=4$$

The equality holds when $\displaystyle \frac{x}{1/x}=\frac{y}{1/y}$, i.e. $x=y$.