An Inequality; $a^2+b^2=1$ $a,b$ are tho real numbers such that  $a^2+b^2=1$.
To prove that ;
$$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{ab+1}\geq\dfrac{3}{1+\cfrac{(a+b)^2}{4}}$$
When I first saw this question, I thought of applying Titu's Lemma, to get
$$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{ab+1}\geq\dfrac{9}{1+1+1+a^2+b^2+ab}=\dfrac{9}{3+(a+b)^2-ab}OR=\dfrac{9}{4+ab}$$
Now, from hereI am confused, how to proceed. Can anybody kindly help me  over this problem ?

Your partial attempt above using Titu's Lemma cannot succeed because if I put $a = \sqrt{3}/2$ and $b = 1/2$, then $9/(4 + ab) < 3 / (1 + (a+b)^2/4)$. 
I would also go with the trigonometric solution above though some care is needed to show that the expression above is positive when one of $a$ or $b$ is negative.
