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For $x$, $y$ $\in R^+$, prove that $$\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}$$

Could someone please help me with this inequality problem? I have tried to use the AM-GM inequality but I must be doing something wrong. I think it can be solved with the AM-GM but I can’t solve it. Thanks in advance for your help.

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What do you get when you cross multiply to clear out the denominators? –  Calvin Lin Apr 25 '13 at 20:27
    
Got it. Thanks. –  user59670 Apr 25 '13 at 20:36
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4 Answers 4

up vote 1 down vote accepted

Here is a solution with AM-GM:

$$\frac{1}{x}+\frac{1}{y} \geq \frac{2}{\sqrt{xy}}$$ $$x+y \geq 2 \sqrt{xy} \Rightarrow \frac{1}{\sqrt{xy}} \geq \frac{2}{x+y}\Rightarrow \frac{2}{\sqrt{xy}} \geq \frac{4}{x+y}$$

Also you can note that

$$(x+y)(\frac{1}{x}+\frac{1}{y}) \geq 4$$ is just Cauchy-Schwartz.

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Brilliant thanks. I guess I need to practice more to get better at these techniques. –  user59670 Apr 25 '13 at 20:40
    
@user59670 BTW if you divide by $2$ your inequality becomes the AM-HM inequality. –  N. S. Apr 25 '13 at 20:42
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Hint: Your inequality is equivalent to $(x-y)^2\ge 0$, which is obviously true.

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Without words:

$$\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\iff \frac{x+y}{xy}\ge \frac{4}{x+y}\iff (x+y)^2\ge4xy\iff (x-y)^2\ge 0$$

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You wrote "Without words:" –  Lord Soth Apr 25 '13 at 20:37
    
:) Well, you can add "after these last ones" if you will. –  DonAntonio Apr 25 '13 at 20:38
    
Your last equivalence could also be $\Leftrightarrow \frac{x+y}{2}\geq \sqrt{xy}$... Of course the one you added is cleaner. –  N. S. Apr 25 '13 at 20:47
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You can also use Chebyshev's inequality because if $x\geq y$ then $\frac{1}{x} \leq \frac{1}{y}$. So, we have: $$(x+y)(\frac{1}{x}+\frac{1}{y})\geq2(x\cdot\frac{1}{x}+y\cdot\frac{1}{y})=4$$

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