Given $a+b+c=3$ .Prove $ \sum \limits_{cyc} \frac {1}{a^2+b^2+2} \le \frac 34$ Yesterday I found this on the Internet:
Give 3 non-negative numbers $a,b,c$ that $a+b+c=3$. Prove
$$ \sum _{cyc} \frac  {1}{a^2+b^2+2} \le \frac 34 $$ 
I have tried to solve this using AM-GM:
From AM-GM we got: 
$$a^2+b^2+2\ge2(a+b)$$
$$\Rightarrow \frac{1}{a^2+b^2+2} \le \frac12\frac1{a+b}$$
$$\Rightarrow \sum _{cyc} \frac  {1}{a^2+b^2+2} \le \frac12\sum_{cyc}\frac{1}{a+b}$$
We have to prove $$\frac12\sum_{cyc}\frac{1}{a+b}\le\frac34$$
or  $$\sum_{cyc}\frac{1}{a+b}\le\frac32$$
However the above statement seem to be false. If $a = 0 , b = 1, c = 2$:
$$\sum_{cyc}\frac{1}{a+b} = 1+\frac13+\frac12=\frac{11}6\gt \frac32$$
Anyone know the solutions?

It's suffice to show the following inequality $$ \sum\limits_{sic}{\frac{a^2+b^2}{a^2+b^2+2}} \ge \frac{3}{2} $$ By using Cauchy, we have $$ LHS \ge \frac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2\left(a^2+b^2+c^2\right)+6} \ge \frac{\sqrt{3\left(a^2b^2+b^2c^2+c^2a^2\right)}+2\left(a^2+b^2+c^2\right)}{\left(a^2+b^2+c^2\right)+3} $$ Notice that $ \sqrt{3\left(a^2b^2+b^2c^2+c^2a^2\right)} \ge ab+bc+ca $, thus, the last term is greater than $$ \frac{2\left(a^2+b^2+c^2\right)+\left(ab+bc+ca\right)}{\left(a^2+b^2+c^2\right)+3} = \frac{3}{2} $$ The conclusion follows. Note that $ a+b+c=3 $ implies $ a^2+b^2+c^2+2ab+2bc+2ca=9 $.
