<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] 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 $5x^2+4 \ge 4\sqrt5 |x|> 6|x|$.
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