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Prove that $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\gt1$, where a, b, c >0.

I tried, $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} = \frac{a}{\sqrt{a}\sqrt{a+b}}+\frac{b}{\sqrt{b}\sqrt{b+c}}+\frac{c}{\sqrt{c}\sqrt{c+a}} \ge \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\sqrt{a}\sqrt{a+b}+\sqrt{b}\sqrt{b+c}+\sqrt{c}\sqrt{c+a}}$.

Then $\sqrt{a}\sqrt{a+b} \le \sqrt{2}/2 (2a+(a+b))$.

This only gave me $\sqrt{2}/2 + \sqrt{2}(\frac{ab+bc+ca}{a+b+c})$.

renmom
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3 Answers3

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Hint: $\dfrac{a}{a+b} > \dfrac{a}{a+b+c}$

DeepSea
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    I think the following is better: $\sqrt{\frac{a}{a+b}}>\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}.$ – Michael Rozenberg Oct 30 '20 at 04:30
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    @Michael Rosenberg: Sure is. I leave it for OP to fill – DeepSea Oct 30 '20 at 04:32
  • Got it. $a+b\lt(\sqrt(a)+\sqrt(b)+\sqrt(c))^2$. Much thanks. – renmom Oct 30 '20 at 04:48
  • @lilige Show your attempts and I'll show another solutions. The same words I can say about the following your topic. https://math.stackexchange.com/questions/3887062/ – Michael Rozenberg Oct 30 '20 at 04:50
  • Michael, I assumed I had to use Cauchy. So I tried times (√) top and bottom, apply Titu's, then denominator sqrt(a(a+b)) using AM-GM. That didn't get me "1". Looking forward to your "another solution". – renmom Oct 30 '20 at 04:55
  • @lilige Add it in your open post. Also, show, how exactly you made these things. – Michael Rozenberg Oct 30 '20 at 04:57
  • Michael, I added to my initial post. – renmom Oct 30 '20 at 05:17
  • @lilige I posted. There are another solutions of your problem. I see someone wants to close your beautiful topic. We'll wait. If it not will happen, I'll post these solutions. – Michael Rozenberg Oct 30 '20 at 07:28
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    @MichaelRozenberg: I voted to close as a duplicate because the same inequality has been solved before (in fact by you). That has nothing to do with “beauty,” even the best question can be a duplicate. It would be better if you voted to close (and extend your older answer, if appropriate) instead of posting two more answers here. – Martin R Oct 30 '20 at 07:45
  • Thank you, Michael. It's fine to close it to clean things up. I'm new here - I'll practice "vote to close" myself next time. – renmom Oct 30 '20 at 20:06
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Another way. $$\sum_{cyc}\sqrt{\frac{a}{a+b}}=\sum_{cyc}\frac{\sqrt{a}}{\sqrt{a+b}}>\sum_{cyc}\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}>\sum_{cyc}\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1.$$

Michael Rozenberg
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We can use also your idea, but in another writing: $$\sum_{cyc}\sqrt{\frac{a}{a+b}}=\sum_{cyc}\frac{\sqrt{a(a+b)}}{a+b}>\sum_{cyc}\frac{\sqrt{a\cdot a}}{a+b}>\sum_{cyc}\frac{a}{a+b+c}=1.$$

Michael Rozenberg
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  • Right! Since the inequality is not "tight", I could just throw away some terms in numerator or add positive terms in the denominator, each ends up different way to solve. Very nice. And thanks a lot. – renmom Oct 30 '20 at 20:18