<s>[INST] 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 ?
 [/INST] Let $a = \cos(t)$, $b = \sin(t)$.  Then (using Maple) the inequality simplifies to
$$ {\frac { \left( 1 + \dfrac32\,\sin \left( 2\,t \right) +\dfrac54\,  \sin^2
 \left( 2\,t \right)    \right)  \left( \cos \left( t
 \right) -\sin \left( t \right)  \right) ^{2}}{ \left(   \cos^2
 \left( t   \right) +1 \right)  \left(   \sin^2 \left( 
t \right)  +1 \right)  \left( \cos \left( t \right) \sin
 \left( t \right) +1 \right)  \left((\cos(t)+\sin(t))^2+4\right)   }} \ge 0 $$
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