<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] 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".
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