Find the minimum value of $$ \frac{a}{1+b} + \frac{b}{1+a} + (1-a)(1-b)$$

when $a,b$ is real numbers and satisfying $a,b \in(0,1)$.


closed as off-topic by Did, Claude Leibovici, mechanodroid, John B, Arnaud D. Oct 9 '17 at 9:04

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  • $\begingroup$ try $a=x,b=y$ and then minimize for $y$? $\endgroup$ – The Dead Legend Sep 30 '17 at 11:34

Let $a+b=2u$ and $ab=v^2$, where $0<v<1$.

Hence, $u\geq v$ and $$f(u)=\frac{a}{1+b} + \frac{b}{1+a} + (1-a)(1-b)=$$ $$=\frac{a+b+a^2+b^2+(1-a^2)(1-b^2)}{(1+a)(1+b)}=$$ $$=\frac{a+b+1-a^2b^2}{1+a+b+ab}=\frac{2u+1-v^4}{1+2u+v^2}.$$

But, $$f'(u)=\frac{2v^2(1-v^2)}{(1+2u+v^2)^2}>0,$$ which says that $f$ increases, which says that $f$ gets a minimal value for a minimal of $u$,

which happens for $u=v$ or for $a=b$ and the rest is smooth:

Let $g(a)=\frac{2a}{1+a}+(1-a)^2.$

Thus, $$g'(a)=\frac{2}{(1+a)^2}+2a-2=\frac{2a(a^2+a-1)}{(1+a)^2},$$ which gives $a_{min}=\frac{\sqrt5-1}{2}$ and $$g\left(a_{min}\right)=\frac{13-5\sqrt5}{2}.$$ Done!


$\dfrac{\partial \left(\dfrac{a}{b+1}+(1-a) (1-b)+\dfrac{b}{a+1}\right)}{\partial a}=0$

$\dfrac{\partial \left(\dfrac{a}{b+1}+(1-a) (1-b)+\dfrac{b}{a+1}\right)}{\partial b}=0$

That is


which simplifies to

$a^2 b^2+2 a b^2-b=0;\;a^2 b^2+2 a^2 b-a=0$

from the first equation we get

$b \left(a^2 b+2 a b-1\right)=0 \to b_1=0;\;b_2=\dfrac{1}{a (a+2)}\quad(*)$

from the second

$a \left(a b^2+2 a b-1\right)\to a_1=0;\;a b^2+2 a b-1=0$

plug the $b_2$ expression in the last equation

$\dfrac{1}{a (a+2)^2}+\dfrac{2}{a+2}-1=0$

$a^3+2 a^2-1=0\to (1 + a) (-1 + a + a^2)=0\to a_2=-1,a_{34}=\dfrac{-1\pm\sqrt{5}}{2} $

as we are interested in positive values, we have $a=\dfrac{\sqrt{5}-1}{2} $

and $b$ plugging in $(*)$

$b=\dfrac{2}{\left(\sqrt{5}-1\right) \left(\frac{1}{2} \left(\sqrt{5}-1\right)+2\right)}=\dfrac{\sqrt{5}-1}{2}$

So for $a=b=\dfrac{\sqrt{5}-1}{2}$ the given expression becomes

$\frac{1}{2} \left(13-5 \sqrt{5}\right)\approx 0.91$

To be sure that this is a minimum we need to do the second derivative test

Write the Hessian matrix

$H(a,b)=\left( \begin{array}{ll} \dfrac{2 b}{(a+1)^3} & -\dfrac{1}{(b+1)^2}+1-\dfrac{1}{(a+1)^2} \\ -\dfrac{1}{(b+1)^2}+1-\dfrac{1}{(a+1)^2} & \dfrac{2 a}{(b+1)^3} \\ \end{array} \right)$

Compute the determinant at the critic point

$\det H(a,b)=\dfrac{a b \left(a \left(-(a+2)^2 b^3-4 (a+2)^2 b^2-2 (2 a (a+4)+5) b+8\right)+8 b+12\right)-1}{(a+1)^4 (b+1)^4}$

$\det H(\dfrac{\sqrt{5}-1}{2},\dfrac{\sqrt{5}-1}{2})=85-38 \sqrt{5}\approx 0.03>0$

as second derivative $ \dfrac{2 b}{(a+1)^3} $ at the critical point is $7-3 \sqrt{5}\approx 0.29>0$

we have that the point $a=b=\dfrac{\sqrt{5}-1}{2}$ is a local minimum.

hope this helps


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