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


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.
