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

 A: For the left hand side it suffices to let $x = 1$, $y = a$, $z = ab$ and then it is true by Nesbitt's inequality in variables $x,y,z$.
For the right inequality, consider $f(a,b) = \frac{1}{a + ab} + \frac{a}{1 + ab} + \frac{ab}{1 + a}$, now note that $$\frac{\partial^2 f}{\partial b^2}=\frac{2a^3}{(ab+1)^3}+\frac{2a^2}{(ab+a)^3}> 0$$ So $f$ is convex in $b$, that means $f(b)$ attains maxima at the endpoints of its domain, that means it suffices to prove the inequality for $b \in \left\{\frac{1}{2},2\right\}$. For $b = 2$ the inequality is equivalent to $$\frac{(a-2)\left(a-\frac{1}{4}\right)\left(a+\frac{5}{9}\right)}{a(a+1)\left(a+\frac{1}{2}\right)}\leq 0$$ which is true on given interval. For $b = \frac{1}{2}$ the inequality is equivalent to $$\frac{(a-4)\left(a-\frac{1}{2}\right)\left(a+\frac{10}{9}\right)}{a(a+1)(a+2)}\leq 0$$ which is true on given interval. Hence the inequality is proved.
EDIT: Your idea for proving left hand side inequality can indeed be used, we have $$\frac{1}{a + ab} + \frac{a}{1 + ab} + \frac{ab}{1 + a} - \frac{3}{2} = \frac{1}{2} \cdot \left(\frac{(1 - a)^2}{(a + ab)(1 + ab)} + \frac{(a - ab)^2}{(ab + 1)(a + 1)} + \frac{(ab - 1)^2}{(1 + a)(a + ab)}\right)$$ and this last expression is indeed non-negative.
EDIT #2: Changed tho word "extrema" to just "maxima".
A: 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.
A: 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:
$$\left[ \begin {array}{cc} 2\,{\frac { \left( 1+b \right) ^{2}}{
 \left( a+ab \right) ^{3}}}-2\,{\frac {b}{ \left( 1+ab \right) ^{2}}}+
2\,{\frac {{b}^{2}a}{ \left( 1+ab \right) ^{3}}}-2\,{\frac {b}{
 \left( 1+a \right) ^{2}}}+2\,{\frac {ab}{ \left( 1+a \right) ^{3}}}&2
\,{\frac {a \left( 1+b \right) }{ \left( a+ab \right) ^{3}}}- \left( a
+ab \right) ^{-2}-2\,{\frac {a}{ \left( 1+ab \right) ^{2}}}+2\,{\frac 
{{a}^{2}b}{ \left( 1+ab \right) ^{3}}}+ \left( 1+a \right) ^{-1}-{
\frac {a}{ \left( 1+a \right) ^{2}}}\\ 2\,{\frac {a
 \left( 1+b \right) }{ \left( a+ab \right) ^{3}}}- \left( a+ab
 \right) ^{-2}-2\,{\frac {a}{ \left( 1+ab \right) ^{2}}}+2\,{\frac {{a
}^{2}b}{ \left( 1+ab \right) ^{3}}}+ \left( 1+a \right) ^{-1}-{\frac {
a}{ \left( 1+a \right) ^{2}}}&2\,{\frac {{a}^{2}}{ \left( a+ab
 \right) ^{3}}}+2\,{\frac {{a}^{3}}{ \left( 1+ab \right) ^{3}}}
\end {array} \right] $$
A: Remarks: Here is a proof  without using calculus.
@Calvin Lin gave a nice proof of the left inequality.
Here we only prove the right inequality.

Let $c = 1/a \in [1/2, 2]$. We have
$$
 \frac{1}{a + ab} + \frac{a}{1 + ab} + \frac{ab}{1 + a} =
 \frac{c}{1+b} + \frac{1}{b+c} + \frac{b}{1+c}.
$$
It suffices to prove that, for all $b, c \in [1/2, 2]$,
$$\frac{c}{1+b} + \frac{1}{b+c} + \frac{b}{1+c} \le \frac{19}{10}.$$
(1) If $b + c \le 5/2$, letting $x = b + c \in [1, 5/2]$, we have
\begin{align*}
 \frac{c}{1+b} + \frac{1}{b+c} + \frac{b}{1+c}
 &= \frac{(b + c)^2 + b + c - 2bc}{1 + b + c + bc} + \frac{1}{b+c}\\[5pt]
 &\le \frac{(b + c)^2 + b + c - 2\cdot \frac12(b + c - \frac12)}{1 + b + c + \frac12(b + c - \frac12)} + \frac{1}{b+c} \tag{1}\\[5pt]
&= \frac{4x^2 + 2}{3 + 6x} + \frac{1}{x}\\[5pt]
 &\le  \frac{19}{10}
\end{align*}
where in (1) we have used $bc - \frac12(b + c - \frac12) = \frac14(2b - 1)(2c - 1) \ge 0$, and the last inequality follows from
$$\frac{19}{10} - \frac{4x^2 + 2}{3 + 6x} - \frac{1}{x}
= \frac{(5x+2)(5-2x)(4x-3)}{30x(1+2x)} \ge 0.$$
(2) If $b + c > 5/2$, letting $x = b + c \in (5/2, 4]$, we have
\begin{align*}
 \frac{c}{1+b} + \frac{1}{b+c} + \frac{b}{1+c}
 &= \frac{(b + c)^2 + b + c - 2bc}{1 + b + c + bc} + \frac{1}{b+c}\\[5pt]
 &\le \frac{(b + c)^2 + b + c - 2\cdot 2(b + c - 2)}{1 + b + c + 2(b + c - 2)} + \frac{1}{b+c} \tag{2}\\[5pt]
  &= \frac{x^2 - 3x + 8}{3x - 3} + \frac{1}{x}\\[5pt]
 &\le \frac{19}{10}
\end{align*}
where in (2) we have used $bc - 2(b + c - 2) = (2 - b)(2 - c) \ge 0$, and the last inequality follows from
$$\frac{19}{10} - \frac{x^2 - 3x + 8}{3x - 3} - \frac{1}{x} = \frac{(6-x)(2x-5)(5x-1)}{30x(x-1)} \ge 0.$$
We are done.


Also, the original inequality can be proved using some substitutions.
Note that
$$a \in [1/2, 2)
\iff a = \frac12 \cdot \frac{1}{1+s} + 2\cdot \frac{s}{1 + s}, \quad s \ge 0$$
and
$$b \in [1/2, 2)
\iff b = \frac12 \cdot \frac{1}{1+t} + 2\cdot \frac{t}{1 + t}, \quad t \ge 0.$$
If $a \in [1/2, 2)$ and $b \in [1/2,2)$,
we have
$$\frac{1}{a + ab} + \frac{a}{1 + ab} + \frac{ab}{1 + a}
= \frac{f(s, t)}{g(s, t)}$$
where $f(s, t)$ and $g(s, t)$
are both polynomials with non-negative coefficients.
The desired result follows.
It remains to prove the case
$a = 2$ or $b = 2$. It is easy.
We are done.
