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.

$\large \dfrac{a}{b + 1} + \dfrac{b}{a + 1} + \dfrac{1}{a + b}$
Taking $a+b = c$ for some c; $0<c\leq 1$, and expressing $b=c-a$; the above expression becomes (after modification) 
$-2 + \{2(c+1) +(c^2+c)\}(\dfrac{1}{-a^2+ac+(c+1)} ) +1/c$ 
here only variable is a ; c is assumed constant. 
Differentiating it and equating it to zero we get
$0+\{2(c+1) +(c^2+c)\}(-\dfrac{-2a+c}{(-a^2+ac+(c+1))^2} ) +0  = 0$
From this we get $a = c/2 = b$, using $a=b$ from here we get
$\dfrac{2a}{a+1}+\dfrac{1}{2a} = \dfrac{4a^2+a+1}{2a^2+2a} = 2 - \dfrac{3}{2a}+\dfrac{2}{a^2+a}$ again differentiatig wrt a and setting it to zero we get
$3a^2-2a-1 = 0$, from this we get $a = 1,-1/3$,
Now let $a = 1-d$ then $\dfrac{2a}{a+1}+\dfrac{1}{2a} = \dfrac{2(1-d)}{2-d}+\dfrac{1}{2(1-d)} = 2 +\dfrac{3d-2}{2(1-d)(2-d)}$  it is very clear from here that above function is strictly increasing from d = 0 to d = 1; or strictly decreasing from a = 0 to a = 1, but the maximum value of "a" possible here is $a=b = 1/2$ 
