Proving an inequality given $\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}\le 1$ Given that $a,b,c > 0$ are real numbers such that $$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}\le 1,$$ prove that $$\frac{1}{b+c+1}+\frac{1}{c+a+1}+\frac{1}{a+b+1}\ge 1.$$

I first rewrote $$\frac{1}{a+b+1} = 1 - \frac{a+b}{a+b+1},$$ so the second inequality can be rewritten as $$\frac{b+c}{b+c+1} + \frac{c+a}{c+a+1} + \frac{a+b}{a+b+1} \le 2.$$ Cauchy-Schwarz gives us $$\sum \frac{a+b}{a+b+1} \geq \frac{(\sum \sqrt{a+b})^2}{\sum a+ b+ 1}.$$ That can be rewritten as $$\frac{2(a+b+c) + 2\sum \sqrt{(a+b)(a+c)}}{2(a+b+c) + 3},$$ which is greater than or equal to $$\frac{2(a+b+c) + 2 \sum(a + \sqrt{bc})}{2(a+b+c) + 3} = \frac{4(a+b+c) + 2 \sum \sqrt{bc}}{2(a+b+c) + 3} \geq 2,$$ which is the opposite of what I want. Additionally, I'm unsure of how to proceed from here.

Since the two inequalities are completely symmetric in $a,b,c$. WLOG, we only need to study the case  $a \ge b \ge c$.
Let $\Lambda = a + b + c + 1$. The two inequalities can be rewritten
as
$$\sum_{cyc}\frac{a}{\Lambda -a } \stackrel{def}{=}\frac{a}{b+c+1} + \frac{b}{c+a+1} + \frac{c}{a+b+1} \le 1\\
\sum_{cyc}\frac{1}{\Lambda-a}\stackrel{def}{=}\frac{1}{b+c+1}
+ \frac{1}{c+a+1} + \frac{1}{a+b+1} \stackrel{?}{\ge} 1
$$
Notice for $x \in (0,\Lambda)$, the map $x \mapsto \frac{1}{\Lambda - x}$ is increasing. We have
$$a \ge b \ge c \quad\implies\quad \frac{1}{\Lambda - a} \ge \frac{1}{\Lambda - b} \ge \frac{1}{\Lambda - c}$$
By Rearrangement inequality, we have
$$\begin{align}
1 \ge \sum_{cyc}\frac{a}{\Lambda-a} \ge \sum_{cyc}\frac{b}{\Lambda-a}
= \frac{b}{\Lambda-a} +\frac{c}{\Lambda-b} + \frac{a}{\Lambda-c}
\\
1 \ge \sum_{cyc}\frac{a}{\Lambda-a} \ge \sum_{cyc}\frac{c}{\Lambda-a}
= \frac{c}{\Lambda-a} +\frac{a}{\Lambda-b} + \frac{b}{\Lambda-c}
\end{align}
$$
From these, using the decomposition you have, one obtain:
$$\sum_{cyc}\frac{1}{\Lambda-a}
= \sum_{cyc}\left(1 - \frac{b+c}{\Lambda-a}\right) = 3 - 
\sum_{cyc}\frac{b+c}{\Lambda - a} 
\ge 3 - (1+1) = 1
$$
