# How prove this inequality $\frac{3}{2}\le\frac{1}{a+ab}+\frac{a}{1+ab}+\frac{ab}{1+a}\le\frac{19}{10}$

let $a,b\in [\dfrac{1}{2},2]$,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 from this:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=487281&p=2730329#p2730329

Thank you everyone.

I have see this same problem

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oh,sory,I have edit – math110 Jul 17 '13 at 6:15

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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nice,It's very nice+1 – math110 Jul 21 '13 at 5:37

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.

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Thank you ,I have edit, can you prove Right?Thank you – math110 Jul 18 '13 at 5:37
@math110 I'm aware of the new inequality that you posted. In this case, the $x, y, z$ are much different ranges. I'm not certain how to immediately proceed. You can try and modify it? – Calvin Lin Jul 18 '13 at 15:15

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: $$\left[ \begin {array}{cc} 2\,{\frac { \left( 1+b \right) ^{2}}{ \left( a+ab \right) ^{3}}}-2\,{\frac {b}{ \left( 1+ab \right) ^{2}}}+ 2\,{\frac {{b}^{2}a}{ \left( 1+ab \right) ^{3}}}-2\,{\frac {b}{ \left( 1+a \right) ^{2}}}+2\,{\frac {ab}{ \left( 1+a \right) ^{3}}}&2 \,{\frac {a \left( 1+b \right) }{ \left( a+ab \right) ^{3}}}- \left( a +ab \right) ^{-2}-2\,{\frac {a}{ \left( 1+ab \right) ^{2}}}+2\,{\frac {{a}^{2}b}{ \left( 1+ab \right) ^{3}}}+ \left( 1+a \right) ^{-1}-{ \frac {a}{ \left( 1+a \right) ^{2}}}\\ 2\,{\frac {a \left( 1+b \right) }{ \left( a+ab \right) ^{3}}}- \left( a+ab \right) ^{-2}-2\,{\frac {a}{ \left( 1+ab \right) ^{2}}}+2\,{\frac {{a }^{2}b}{ \left( 1+ab \right) ^{3}}}+ \left( 1+a \right) ^{-1}-{\frac { a}{ \left( 1+a \right) ^{2}}}&2\,{\frac {{a}^{2}}{ \left( a+ab \right) ^{3}}}+2\,{\frac {{a}^{3}}{ \left( 1+ab \right) ^{3}}} \end {array} \right]$$

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