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 ?
 A: This is more of a step-by-step version of the trigonometric soln already posted.  With $a=\cos t, \; b = \sin t$, the inequality is 
$$\frac1{\sin^2t+1}+\frac1{\cos^2t+1}+\frac1{\frac12 \sin 2t+1} \ge \frac{12}{5+\sin 2t}$$
$$LHS = \frac{3}{\frac14\sin^22t + 2}+\frac1{\frac12\sin 2t+1}$$
so with $x = \sin 2t\in [-1, 1]$, the inequality can be written as
$$\frac{12}{8+x^2}+\frac2{2+x}\ge \frac{12}{5+x} \iff (1-x)(4+6x+5x^2) \ge 0$$
which is obvious as $5x^2+4 \ge 4\sqrt5 |x|> 6|x|$.
A: Let $a = \cos(t)$, $b = \sin(t)$.  Then (using Maple) the inequality simplifies to
$$ {\frac { \left( 1 + \dfrac32\,\sin \left( 2\,t \right) +\dfrac54\,  \sin^2
 \left( 2\,t \right)    \right)  \left( \cos \left( t
 \right) -\sin \left( t \right)  \right) ^{2}}{ \left(   \cos^2
 \left( t   \right) +1 \right)  \left(   \sin^2 \left( 
t \right)  +1 \right)  \left( \cos \left( t \right) \sin
 \left( t \right) +1 \right)  \left((\cos(t)+\sin(t))^2+4\right)   }} \ge 0 $$
A: 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.
A: 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!
