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Could anyone advise me on how to prove this inequality:

$$\dfrac{a+b}{\sqrt{c}}+\dfrac{a+c}{\sqrt{b}}+ \dfrac{b+c}{\sqrt{a}} \geq 2(\sqrt{a} + \sqrt{b} +\sqrt{c}),$$

where $a,b,c $ are any positive real numbers.

Do I use the AM-GM inequality somewhere?

Thank you.

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  • $\begingroup$ Sorry, I made a typo error. It should be $\geq$. $\endgroup$ – Alexy Vincenzo Nov 4 '15 at 1:01
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From AM-GM you have

$$ \frac{a}{\sqrt{b}} +\frac{a}{\sqrt{b}} +\frac{b}{\sqrt{a}} \geq 3 \sqrt{a} $$ and $$ \frac{a}{\sqrt{c}} +\frac{a}{\sqrt{c}} +\frac{c}{\sqrt{a}} \geq 3 \sqrt{a} $$ Doing this for the other variables and summing respectively you get $$3\left ( \dfrac{a+b}{\sqrt{c}}+\dfrac{a+c}{\sqrt{b}}+ \dfrac{b+c}{\sqrt{a}} \right) \geq 6( \sqrt{a} + \sqrt{b} +\sqrt{c} )$$ Simplifying gives the desired inequality.

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