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)}$

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

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


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!

  • 1
    $\begingroup$ On the second line, shouldn't it be $8(a+b+c)^3$ rather than $8(a+b+c)^2$ ? $\endgroup$ – Ewan Delanoy Aug 12 '17 at 12:41
  • $\begingroup$ Thank you dear @Ewan Delanoy! I fixed my post. $\endgroup$ – Michael Rozenberg Aug 12 '17 at 12:43
  • $\begingroup$ Thank you for your help :) $\endgroup$ – carat Aug 12 '17 at 13:24
  • $\begingroup$ @carat You are welcome! $\endgroup$ – Michael Rozenberg Aug 12 '17 at 13:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.