Prove that $\frac{3}{2}\le\frac{1}{a+ab}+\frac{a}{1+ab}+\frac{ab}{1+a}\le\frac{19}{10}$ for $a, b \in [1/2, 2]$ 
Let
$a,b\in [\frac{1}{2},2]$. Prove that
$$\dfrac{3}{2}\le\dfrac{1}{a+ab}+\dfrac{a}{1+ab}+\dfrac{ab}{1+a}\le\dfrac{19}{10}.$$

my idea:
$$\dfrac{1}{a+ab}+\dfrac{a}{1+ab}+\dfrac{ab}{1+a}-\dfrac{3}{2}\ge 0?$$
and this problem is from《Mathematics Studying》(2012.7).
See: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=487281&p=2730329#p2730329
Thank you everyone.
I have see this  same problem


For the inequality on the left, add 1 to each term. We WTS
$$ \frac{9}{2} \leq (1 + a + ab) \left( \frac{1}{a+ab} + \frac{1}{1+ab} + \frac{1}{1+a} \right). $$
This is equivalent to the following Cauchy Schwarz inequality
$$9 \leq (x+y+z) \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)$$
with $x = 1+a, y = a+ab, z = ab+1$.
The inequality on the right is more interesting.
