<s>[INST] Find the minimum value of $\frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$ where $a, b > 0$ and $a + b \le 1$. 
$a$ and $b$ are two positives such that $a + b \le 1$. Find the minimum value of  $$\large \frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$$

We have that $\dfrac{a}{b + 1} + \dfrac{b}{a + 1} = \dfrac{a^2 + b^2 + a + b}{ab + a + b + 1} \ge \dfrac{\dfrac{(a + b)^2}{2} + (a + b)}{\dfrac{(a + b)^2}{4} + 2(a + b)} = \dfrac{2(a + b) + 4}{(a + b) + 8}$
$\implies \dfrac{a}{b + 1} + \dfrac{b}{a + 1} + \dfrac{1}{a + b} \ge \dfrac{2(a + b) + 4}{(a + b) + 8} + \dfrac{1}{a + b}$
Let $x = 1 - (a + b) \implies x \ge 0$.
Thus $\dfrac{a}{b + 1} + \dfrac{b}{a + 1} + \dfrac{1}{a + b} \ge \dfrac{2(1 - x) + 4}{(1 - x) + 8} + \dfrac{1}{1 - x} = \dfrac{2x^2 - 9x + 15}{x^2 - 10x + 9}$
$= \dfrac{x(x + 23)}{3(9 - x)(1 - x)} + \dfrac{5}{3} \ge \dfrac{5}{3}$
The equality sign happens when $a = b$ and $x = 0$ or $a + b = 1 \implies a = b = \dfrac{1}{2}$.
I want to ask if the above solution is correct and if there are any other solutions that are more rational and reasonable.
 [/INST] Let $y=a+b$. Your expression is the same as $$A=\frac{y+1}{b+1}-1+\frac{y+1}{a+1}-1+\frac{1}{y}=\frac{(y+1)(y+2)}{ab+y+1}+\frac{1}{y}-2$$ 
Now, by  $GM\leq AM$ we have $ab\leq (y/2)^2$. And therefore
$$A\geq \frac{(y+1)(y+2)}{(y/2)^2+y+1}+\frac{1}{y}-2=\frac{(y+1)(y+2)}{(y/2+1)^2}+\frac{1}{y}-2=\frac{4(y+1)}{y+2}+\frac{1}{y}-2$$
Using that $\frac{4(y+1)}{y+2}=4-\frac{4}{y+2}$. We conclude
$$A\geq f(y)=2+\frac{1}{y}-\frac{4}{y+2}$$
Since $f'(y)=\ln(\frac{y}{(y+2)^4})\leq \ln(1)=0$ when $0<y\leq 1$ we conclude that $f(y)$ is decreasing in this interval, so
$$A\geq f(y)\geq f(1)=3/5.$$
The equality happens when $a=b$ and $a+b=y=1$, i.e. when $a=b=1/2$.
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