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Let $a$, $b$, $c>0$ and $ab+bc+ca=3$, prove that $\sum_{cyc}{\frac{a^2+b^2}{a+b+1}}\geq \frac{6abc+6}{a+b+c+3}$.

I have tried using Holder $\sum_{cyc}{\frac{a^2+b^2}{a+b+1}}\geq \frac{2(a+b+c)^2}{\sum_{cyc}{a+b+1}}$

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closed as off-topic by Namaste, Jyrki Lahtonen, José Carlos Santos, Leucippus, Xander Henderson Jun 29 '18 at 1:14

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  • $\begingroup$ Sorry for editing so much i am using my phone . $\endgroup$ – mathlearning Jan 19 '18 at 13:48
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Your idea works!

Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$ and $abc=w^3$.

Hence, $u\geq v\geq w$ and by C-S we obtain: $$\sum_{cyc}\frac{a^2+b^2}{a+b+1}\geq\frac{2(a+b+c)^2}{\sum\limits_{cyc}(2a+1)}=\frac{2(a+b+c)^2}{2(a+b+c)+3}=$$ $$=\frac{18u^2}{6u+3v}=\frac{6u^2}{2u+v}.$$ Thus, it's enough to prove that $$\frac{6u^2}{2u+v}\geq\frac{6(abc+1)}{a+b+c+3}$$ and since $w^3\leq v^3$, it's remains to prove that $$\frac{6u^2}{2u+v}\geq\frac{6(v^3+v^3)}{(3u+3v)v}$$ or $$\frac{u^2}{2u+v}\geq\frac{2v^2}{3(u+v)}$$ or $$(u-v)(3u^2+6uv+2v^2)\geq0.$$ Done!

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  • $\begingroup$ So we need to prove that $3u^3+3u^2\geq 2uw^3+2u+vw^3+v$ $\endgroup$ – mathlearning Jan 19 '18 at 15:15
  • $\begingroup$ Yes, but $v=1$ and $w^3\leq v^3$. Thus, we need to prove that $3u^3v+3u^2v^2\geq2uv^3+2uv^3+v^4+v^4,$ which gives the last inequality in my solution. $\endgroup$ – Michael Rozenberg Jan 19 '18 at 15:20
  • $\begingroup$ Thank you. The last part i didn't know how to prove. $\endgroup$ – mathlearning Jan 19 '18 at 15:21
  • $\begingroup$ It's just factoring. it's $3u^3+3uv^2-4uv^2-2v^3\geq0$ or $3u^3-3u^2v+6u^2v-6uv^2+2uv^2-2v^3\geq0$ or $(u-v)(3u^2+6uv+2v^2)\geq0.$ $\endgroup$ – Michael Rozenberg Jan 19 '18 at 15:22

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