Prove that for all positive real numbers $x$ and $y$, $(x+y)(\frac1x+\frac1y)\ge4$.

Question

Prove that for all positive real numbers $x$ and $y$, $(x+y)(\frac1x+\frac1y)\ge4$.

Any help would be appreciated thanks.

Answer 1

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 $a = x/y$.

Answer 2

Hint:

$$(x + y)\left(\frac{1}{x} + \frac{1}{y}\right)\geq 4 \iff (x+y)\left(\frac{1}{x} + \frac{1}{y}\right) xy \geq 4xy$$

Answer 3

Cauchy-Schwarz gives $$(x+y)(\frac1x+\frac1y)= \left( \left(\sqrt{x}\right)^2 + \left(\sqrt{y}\right)^2 \right)\left( \frac{1}{\left(\sqrt{x}\right)^2} + \frac{1}{\left(\sqrt{y}\right)^2} \right)\ge \left(\frac{\sqrt{x}}{\sqrt{x}} + \frac{\sqrt{y}}{\sqrt{y}}\right)^2 = 4$$