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Let $a$, $b$ and $c$ be positive real numbers. Prove that: $$ \displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}} \geq 2 \sqrt{a+b+c}.$$

My attempt :

By Holder inequality,

$ \left(\displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}}\right)^2 \displaystyle\sum_{cyc}(a+2c)\geq \left(\displaystyle\sum_{cyc} \sqrt{a+b}\right)^3$

$ \left(\displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}}\right)^2 \geq \frac{\left(\displaystyle\sum_{cyc} \sqrt{a+b}\right)^3}{\displaystyle\sum_{cyc}(a+2c)}$

$\Leftrightarrow \left(\displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}}\right)^2 \geq \frac{\left(\displaystyle\sum_{cyc} \sqrt{a+b}\right)^3}{3\displaystyle\sum_{cyc}a}$

$\Leftrightarrow \displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}} \geq \sqrt{\frac{\left(\displaystyle\sum_{cyc} \sqrt{a+b}\right)^3}{3\displaystyle\sum_{cyc}a}}$

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1 Answer 1

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Your idea to use Holder was very good!

By Holder $$\left(\sum_{cyc}\frac{a+b}{\sqrt{a+2c}}\right)^2\sum_{cyc}(a+b)(a+2c)\geq8(a+b+c)^3.$$ Thus, it's enough to prove that $$8(a+b+c)^3\geq4(a+b+c)\sum_{cyc}(a+b)(a+2c)$$ or $$2(a+b+c)^2\geq\sum_{cyc}(a+b)(a+2c),$$ which is $$\sum_{cyc}(a-b)^2\geq0.$$ Done!

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    $\begingroup$ On the second line, shouldn't it be $8(a+b+c)^3$ rather than $8(a+b+c)^2$ ? $\endgroup$ Commented Aug 12, 2017 at 12:41
  • $\begingroup$ Thank you dear @Ewan Delanoy! I fixed my post. $\endgroup$ Commented Aug 12, 2017 at 12:43
  • $\begingroup$ Thank you for your help :) $\endgroup$
    – user403160
    Commented Aug 12, 2017 at 13:24
  • $\begingroup$ @carat You are welcome! $\endgroup$ Commented Aug 12, 2017 at 13:52

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