<s>[INST] Find the minimum value of $\frac ab+\frac {b}{a+b+1}+\frac {b+1}{a}$ when $a,b>0$ I was trying to solve this question and I observed that if I use partial derivative then it would be calculating. Moreover, I observed that $\frac ab\times\frac {b}{a+b+1}\times\frac {b+1}{a}=\frac {b+1}{a+b+1}$. So I thought to use AM-GM inequality and got
$3(\frac {b+1}{a+b+1})^\frac13 \leq \frac ab+\frac {b}{a+b+1}+\frac {b+1}{a}$. 
Now the equality holds if $\frac ab=\frac {b}{a+b+1}=\frac {b+1}{a}$.
From here I also got these relations $\frac {b+1}{a+b+1}=\frac a{a+b}$ and $\frac ab=\frac {b}{a+b+1}=\frac {b+1}{a}=\frac {a+b+1}{a+b}$. I am hoping that I will get a finite expression of $(\frac {b+1}{a+b+1})^\frac13 $ and then we are done as the particular values of $a,b$ are not asked.
Help me from here. Am I in the correct way? Do you have any other suggestions?
 [/INST] For $a>0$ and $b>0$ let $1=t$ and $f(t)=\frac{a}{b}+\frac{b}{a+b+t}+\frac{b+t}{a},$ where $t\geq0.$
Thus, $f'(t)=\frac{1}{a}-\frac{b}{(a+b+t)^2}>0,$ which says $$f(t)\geq f(0)=\frac{a}{b}+\frac{b}{a+b}+\frac{b}{a}.$$
Now, let $a=bx$, where $x>0$.
Thus, we need to find $\min\limits_{x>0}g,$ where $$g(x)=x+\frac{1}{x+1}+\frac{1}{x}.$$
But $$g'(x)=1-\frac{1}{(x+1)^2}-\frac{1}{x^2}=\frac{x^4+2x^3-x^2-2x-1}{x^2(x+1)^2}=$$
$$=\frac{(x^2+x-1)^2-2}{x^2(x+1)^2}=\frac{(x^2+x-1-\sqrt2)(x^2+x+\sqrt2-1)}{x^2(x+1)^2},$$ which gives $$x_{min}=\frac{-1+\sqrt{5+4\sqrt2}}{2}.$$
Can you end it now?
I got that the infimum is equal to $$\frac{2\sqrt{10+8\sqrt2}-\sqrt{5+4\sqrt2}-1}{2}=\frac{(2\sqrt2-1)\sqrt{5+4\sqrt2}-1}{2}=$$
$$=\frac{\sqrt{(9-4\sqrt2)(5+4\sqrt2)}-1}{2}=\frac{\sqrt{13+16\sqrt2}-1}{2}\approx2.4844...$$
We can use also the following reasoning.
Let $a=bx$.
Thus, $$\frac{a}{b}+\frac{b}{a+b+1}+\frac{b+1}{a}=x+\frac{b}{bx+b+1}+\frac{b+1}{bx}=$$
$$=x+\frac{1}{x+1+\frac{1}{b}}+\frac{1}{x}+\frac{1}{bx}>x+\frac{1}{x+1}+\frac{1}{x}$$ because
$$x+\frac{1}{x+1+\frac{1}{b}}+\frac{1}{x}+\frac{1}{bx}-\left(x+\frac{1}{x+1}+\frac{1}{x}\right)=$$
$$=\frac{1}{bx}-\left(\frac{1}{x+1}-\frac{1}{x+1+\frac{1}{b}}\right)=\frac{1}{bx}-\frac{\frac{1}{b}}{(x+1)\left(x+1+\frac{1}{b}\right)}>0.$$
Id est, it's enough to find an infimum of $x+\frac{1}{x+1}+\frac{1}{x}$ and the rest is the same.
We see the for $b\rightarrow+\infty$ the expression $\frac{a}{b}+\frac{b}{a+b+1}+\frac{b+1}{a}$ is closed to $x+\frac{1}{x+1}+\frac{1}{x}$, which says that $$\inf_{a>0,b>0}\left(\frac{a}{b}+\frac{b}{a+b+1}+\frac{b+1}{a}\right)=\inf_{x>0}\left(x+\frac{1}{x+1}+\frac{1}{x}\right)=\frac{\sqrt{13+16\sqrt2}-1}{2}.$$
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