# Inequality with $ab+bc+ca=3$ [closed]

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 $\sum_{cyc}{\frac{a^2+b^2}{a+b+1}}\geq \frac{2(a+b+c)^2}{\sum_{cyc}{a+b+1}}$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Jyrki Lahtonen, José Carlos Santos, Leucippus, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.

• Sorry for editing so much i am using my phone . – mathlearning Jan 19 '18 at 13:48

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!
• So we need to prove that $3u^3+3u^2\geq 2uw^3+2u+vw^3+v$ – mathlearning Jan 19 '18 at 15:15
• 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. – Michael Rozenberg Jan 19 '18 at 15:20
• 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 $(u-v)(3u^2+6uv+2v^2)\geq0.$ – Michael Rozenberg Jan 19 '18 at 15:22