<s>[INST] 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

 [/INST] You can also use the Lagrange Function for the optimisation problem
$$\min_{(a,b) \in [\frac{1}{2},2]^2} f(a,b) := \frac{1}{a+ab} + \frac{a}{1+ab} + \frac{ab}{1+a}$$
This will result in an inner minimum at $(a,b) = (1,1), f_{min} = \frac{3}{2}$ and boundary maxima at $(a,b) \in \{ (\frac{1}{2},\frac{1}{2}), (2,2) \}, f_{max} = \frac{19}{10}$ as proposed. The biggest annoyance in this solution is the Hessian Matrix of $f$, wich is very ugly:
$$\left[ \begin {array}{cc} 2\,{\frac { \left( 1+b \right) ^{2}}{
 \left( a+ab \right) ^{3}}}-2\,{\frac {b}{ \left( 1+ab \right) ^{2}}}+
2\,{\frac {{b}^{2}a}{ \left( 1+ab \right) ^{3}}}-2\,{\frac {b}{
 \left( 1+a \right) ^{2}}}+2\,{\frac {ab}{ \left( 1+a \right) ^{3}}}&2
\,{\frac {a \left( 1+b \right) }{ \left( a+ab \right) ^{3}}}- \left( a
+ab \right) ^{-2}-2\,{\frac {a}{ \left( 1+ab \right) ^{2}}}+2\,{\frac 
{{a}^{2}b}{ \left( 1+ab \right) ^{3}}}+ \left( 1+a \right) ^{-1}-{
\frac {a}{ \left( 1+a \right) ^{2}}}\\ 2\,{\frac {a
 \left( 1+b \right) }{ \left( a+ab \right) ^{3}}}- \left( a+ab
 \right) ^{-2}-2\,{\frac {a}{ \left( 1+ab \right) ^{2}}}+2\,{\frac {{a
}^{2}b}{ \left( 1+ab \right) ^{3}}}+ \left( 1+a \right) ^{-1}-{\frac {
a}{ \left( 1+a \right) ^{2}}}&2\,{\frac {{a}^{2}}{ \left( a+ab
 \right) ^{3}}}+2\,{\frac {{a}^{3}}{ \left( 1+ab \right) ^{3}}}
\end {array} \right] $$
</s>