Prove that for all positive real numbers $x$ and $y$, $(x+y)(\frac1x+\frac1y)\ge4$. Prove that for all positive real numbers $x$ and $y$,
$(x+y)(\frac1x+\frac1y)\ge4$.
Any help would be appreciated thanks.
 A: To begin with, notice that
\begin{align*}
(x+y)\left(\frac{1}{x} + \frac{1}{y}\right) = 2 + \frac{x}{y} + \frac{y}{x}
\end{align*}
According to the inequality $AM\geq GM$, we have
\begin{align*}
\frac{x}{y} + \frac{y}{x} \geq 2\sqrt{\frac{x}{y}\cdot\frac{y}{x}} = 2
\end{align*}
from whence we obtain the claimed result.
EDIT
Due to JavaMan's observation, I provide you another approach to solve the second part. Indeed, for any $a\in\textbf{R}_{>0}$, we have
$$(a-1)^{2}\geq 0 \Leftrightarrow a^{2} - 2a + 1 \geq 0 \Leftrightarrow a^{2} + 1 \geq 2a\Leftrightarrow a + \frac{1}{a} \geq 2$$
In the present case, it suffices to take $a = x/y$.
A: Hint:
$$(x + y)\left(\frac{1}{x} + \frac{1}{y}\right)\geq 4 \iff (x+y)\left(\frac{1}{x} + \frac{1}{y}\right) xy \geq 4xy$$
A: Cauchy-Schwarz gives
$$(x+y)(\frac1x+\frac1y)= \left( \left(\sqrt{x}\right)^2 + \left(\sqrt{y}\right)^2 \right)\left( \frac{1}{\left(\sqrt{x}\right)^2} + \frac{1}{\left(\sqrt{y}\right)^2} \right)\ge \left(\frac{\sqrt{x}}{\sqrt{x}} + \frac{\sqrt{y}}{\sqrt{y}}\right)^2 = 4$$
