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Prove that if $$ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c}=2 $$ then $$\sum_{cyc} \sqrt{ab} \geq \frac{3}{2}$$

I tried to use AM-GM inequality and Titu's lemma and was able to show that $a+b+c\geq3$ and $\sum_{cyc}\sqrt{ab}\geq3abc$.

Can someone give a hint on how to continue?

Edit: I apologize for all the typos in the original submission.(I'm new to the website and i don't know if i should delete the original post) This is my work on the probleme: By Titu's lemma we have:

$$ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c} \geq \frac{9}{2a+2b+2c}$$ Then by using the condition we end up with $$ a+b+c \geq \frac{9}{4} $$ by AM-GM we have: $$ \sum_{cyc} \frac{1}{a+b} \leq \sum_{cyc} \frac{1}{2\sqrt{ab}} $$ so $$\sum_{cyc} \frac{1}{\sqrt{ab}} \geq 4 $$ in the other hand by AM-GM we have $$ 3\sum_{cyc} \frac{1}{\sqrt{ab}} \leq (\sum_{cyc} \frac{1}{\sqrt{a}})^{2}$$ using those later results i ended up with $$\sum_{cyc}\sqrt{ab}\geq \sqrt{12abc}$$

roro roro
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  • Consider $a=b=c=\frac34$, then $a+b+c=\frac94<3$. Can you show your working for those results you got? – Macavity Jan 04 '24 at 13:13
  • Smells like Jensen inequality to me. – Mark Jan 04 '24 at 13:19
  • What $\sum \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c}$ means? Did you meant $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c}$? – jjagmath Jan 04 '24 at 13:39
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    The inequality is incorrect, check $a= b=5/4, c = 0$. Then $\sum_{cyc} \frac{1}{a+b} = 2$ but $\sum_{cyc}\sqrt{ab} = \sqrt{ab} = 5/4 < 3/2.$ – V.S.e.H. Jan 04 '24 at 17:18

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