Given is equivalent to $\displaystyle \sum \dfrac{(a+b)^2+ ab+bc+ac}{c^2+2(ab+bc+ca)}\geq3\Leftrightarrow \sum \dfrac{(a+b)^2+ ab+bc+ac}{(c+a)(c+b)+ab+bc+ca}\geq 3$ Set $a+b=x,b+c=y,c+a=z,ab+bc+ac=t$.We have to show that $\displaystyle \sum \dfrac{x^2+t}{yz+t}\geq 3$ By AM-GM we have that $\displaystyle \sum \dfrac{x^2+t}{yz+t}\geq 3\sqrt[3]{\dfrac{\displaystyle\prod(x^2+t)}{\displaystyle\prod(yz+t)}}$ so it suffice to show that $\displaystyle\prod(x^2+t)\geq \prod(yz+t) $ which is true because by Cauchy–Schwarz inequality we have that $\displaystyle \prod(x^2+t)^2=\prod [(x^2+t)(y^2+t)]\geq\prod(xy+t)^2$

parmenides51 wrote: Let $a, b, c$ be real numbers with the property as $ab + bc + ca = 1$. Show that: $$\frac {(a + b) ^ 2 + 1} {c ^ 2 + 2} + \frac {(b + c) ^ 2 + 1} {a ^ 2 + 2} + \frac {(c + a) ^ 2 + 1} {b ^ 2 + 2} \ge 3 $$. See also here

$LHS=\sum{\frac{(a+b)^2}{c^2+2}}+\sum{\frac{1}{c^2+2}} \geq \frac{4(a+b+c)^2}{a^2+b^2+c^2+6}+\frac{9}{a^2+b^2+c^2+6}=\frac{4a^2+4b^2+4c^2+17}{a^2+b^2+c^2+6} \geq 3$ $4a^2+4b^2+4c^2+17 \geq 3a^2+3b^2+3c^2+18$ $a^2+b^2+c^2 \geq 1=ab+bc+ca$ which is true.

this stuff sounds interesting...

$a^2+2=a^2+2*(ab+bc+ca)=(a+b)*(a+c)+ab+bc+ca$ just use some subs and try cauchy(maybe holder also helps) also another method is $(a+b)^2+1=(a+b)*(a+c)+b^2+2ab=(a+b)*(a+c)+b*(b+a)+ab=(b+a)(a+b+c)+ab$

Prod55 wrote: Set $a+b=x,b+c=y,c+a=z,ab+bc+ac=t$.We have to show that $\displaystyle \sum \dfrac{x^2+t}{yz+t}\geq 3$ By AM-GM we have that $\displaystyle \sum \dfrac{x^2+t}{yz+t}\geq 3\sqrt[3]{\dfrac{\displaystyle\prod(x^2+t)}{\displaystyle\prod(yz+t)}}$ The numbers $a$, $b$, $c$ (and hence $x$, $y$, $z$) are not necessarily non-negative, so you cannot apply AM-GM, I think?

$\sum_{cyc} \frac{(a+b)^2+1}{c^2+2}=\sum_{cyc} \frac{(a+b)^2}{c^2+2}+\sum_{cyc} \frac{1^2}{c^2+2}\geq \frac{4(a+b+c)^2}{a^2+b^2+c^2+6}+\frac{9}{a^2+b^2+c^2+6}=\frac{4(a^2+b^2+c^2+2)}{a^2+b^2+c^2+6}+\frac{9}{a^2+b^2+c^2+6}=\frac{4(a^2+b^2+c^2+6)}{a^2+b^2+c^2+6}-\frac{16}{a^2+b^2+c^2+6}+\frac{9}{a^2+b^2+c^2+6}=4-\frac{7}{a^2+b^2+c^2+6}\geq 4-\frac{7}{ab+bc+ca+6}=3$