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.

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.
