# Find the minimum value of $a/(1+b) + b/(1+a) + (1-a)(1-b)$ [closed]

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Claude Leibovici, mechanodroid, John B, Arnaud D.
If this question can be reworded to fit the rules in the help center, please edit the question.

• try $a=x,b=y$ and then minimize for $y$? – 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

$$-\frac{b}{(a+1)^2}+b+\frac{1}{b+1}-1=0;\;-\frac{a}{(b+1)^2}+a+\frac{1}{a+1}-1=0$$

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