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.
 A: 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 $a=b=1/2$.
A: $\large \dfrac{a}{b + 1} + \dfrac{b}{a + 1} + \dfrac{1}{a + b}$
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 $a=b = 1/2$ 
A: Your answer is almost right - you just need to show there exists $a$ and $b$ where $\frac{5}{3}$ is attainable (this happens at $a = b = \frac{1}{2}).
Here is an alternative solution. Let $c = 1+a+b$. The expression becomes $\frac{a}{c-a} + \frac{b}{c-b} + \frac{1}{c-1}$. We claim that for fixed $c$, this is minimised when $a=b$. Let $f_c(x) = \frac{x}{c-x} = \frac{c}{c-x}-1$. This expression can be easily shown to be convex, and so by Jensen's inequality, if we let $d = \frac{a+b}{2}$, we have $\frac{2d}{c-d} \le \frac{a}{c-a} + \frac{b}{c-b}$, proving the claim.
We now need to minimise $\frac{2x}{1+x} + \frac{1}{2x} = 2-\frac{2}{1+x}+\frac{1}{x}$. It is simple to show that this is strictly decreasing on $(0,\frac{1}{2}]$, so the minimum is obtained at $x = \frac{1}{2}$ which gives $\frac{5}{3}$ as required.
