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

Any help would be appreciated thanks.

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  • $\begingroup$ What have you tried? $\endgroup$ – JavaMan Jan 21 at 4:17
  • $\begingroup$ I tried expanding the left side but I do not know how to show that the left sider is also greater than 4, only equal to. $\endgroup$ – macy Jan 21 at 4:19
  • $\begingroup$ Use $AM \geq HM$ $\endgroup$ – Anik Bhowmick Jan 21 at 5:43
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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 $a = x/y$.

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  • $\begingroup$ AM-GM is too much here. Simple algebra will do it. $\endgroup$ – JavaMan Jan 21 at 4:21
  • $\begingroup$ If you take a=x/y, doesn't the rhs still remain as 2? $\endgroup$ – macy Jan 21 at 15:57
  • $\begingroup$ Don't forget to add the previous result: $$(x+y)\left(\frac{x}{y} + \frac{y}{x}\right) = 2 + \frac{x}{y} + \frac{y}{x} \geq 2 + 2 = 4$$ $\endgroup$ – user1337 Jan 21 at 15:58
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Hint:

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

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  • $\begingroup$ Why is it necessary to multiply by xy here? $\endgroup$ – macy Jan 21 at 15:56
  • $\begingroup$ It's not necessary, but it does cancel the fractions. It turns the equality into $(x+y)^2 \geq 4xy$ which is equivalent to $(x-y)^2 \geq 0$, which is obviously true. $\endgroup$ – JavaMan Jan 21 at 21:48
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Cauchy-Schwarz gives $$(x+y)(\frac1x+\frac1y)= \left( \left(\sqrt{x}\right)^2 + \left(\sqrt{y}\right)^2 \right)\left( \frac{1}{\left(\sqrt{x}\right)^2} + \frac{1}{\left(\sqrt{y}\right)^2} \right)\ge \left(\frac{\sqrt{x}}{\sqrt{x}} + \frac{\sqrt{y}}{\sqrt{y}}\right)^2 = 4$$

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