let $a,b>0$,find the minimum of the value $f(a,b)=\frac{a}{b}+\frac{b}{a+b+1}+\frac{b+1}{a}$

Question

let $a,b>0$,find the minimum of the value $$f(a,b)=\dfrac{a}{b}+\dfrac{b}{a+b+1}+\dfrac{b+1}{a}$$

I try $$f'_{a}=\dfrac{1}{b}-\dfrac{b}{(a+b+1)^2}-\dfrac{b+1}{a^2}=0$$ $$f'_{b}=-\dfrac{a}{b^2}+\dfrac{a+1}{(a+b+1)^2}+\dfrac{1}{a}=0$$ it seem not easy find the $a=?b=?$ so How to find the minimum of the value

Answer 1

Let $a=bx$ and $y=1+\frac{1}{b}.$

Thus, $y>1$ and $$\frac{a}{b}+\frac{b}{a+b+1}+\frac{b+1}{a}=x+\frac{1}{x+y}+\frac{y}{x}.$$ But $$\frac{\partial}{\partial y}\left(x+\frac{1}{x+y}+\frac{y}{x}\right)=\frac{1}{x}-\frac{1}{(x+y)^2}>\frac{1}{x}-\frac{1}{(x+1)^2}>0,$$ which says that it's enough to find $$\min_{x>0}\left(x+\frac{1}{x+1}+\frac{1}{x}\right).$$ Let $g(x)=x+\frac{1}{x+1}+\frac{1}{x}.$

Thus, $$g'(x)=1-\frac{1}{(x+1)^2}-\frac{1}{x^2}=\frac{x^4+2x^3-x^2-2x-1}{x^2(x+1)^2}=$$ $$=\frac{(x^2+x-1)^2-2}{x^2(x+1)^2}=\frac{(x^2+x-1-\sqrt2)(x^2+x-1+\sqrt2)}{x^2(x+1)},$$ which gives $$x_{min}=\frac{\sqrt{5+4\sqrt2}-1}{2}.$$ Can you end it now?

The minimum does not exist, but the infimum we can evaluate.

I got $$\inf_{a>0,b>0} f=\frac{(2\sqrt2-1)\sqrt{5+4\sqrt2}-1}{2}\approx2.4844...$$

Answer 2

This is not an answer

I wonder if the function is correct (typo may be).

Using a lot of algebra, the solutions of $f'_a=f'_b=0$ are $$\{a= -1,b= -1\}$$ $$\left\{a= \frac{-1+i \sqrt{3}}{4} ,b= \frac{-1-i \sqrt{3}}{4} \right\}$$ $$\left\{a= \frac{-1-i \sqrt{3}}{4} ,b= \frac{-1+i \sqrt{3}}{4}\right\}$$ None of them satisfies $a>0$, $b>0$.

Using the positivity constraints and 3D plots as well as contour plots, it seems that the minimum is around $2.5$ but, for this specific level, the contour is a very long valley in the $(a,b)$ plane.