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 ?

We need to prove that
$$\frac{1}{2a^2+b^2}+\frac{1}{a^2+2b^2}+\frac{1}{a^2+ab+b^2}\geq\frac{12}{5a^2+2ab+5b^2}$$ or
$$(a-b)^2(a^4+3a^3b+7a^2b^2+3ab^3+b^4)\geq0,$$
which is obvious.
Done!
