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.

Applying Jensens to $ f(x) = \frac{ x} { (a+b+c+1) - x } $, we have $$ 1\geq \sum \frac{a}{b+c+1} = f(a) + f(b) + f(c)  \geq 3 f ( \frac{a+b+c } { 3} ) = 3 \times \frac{  a + b + c } { 2a + 2b + 2c + 3 } \Rightarrow a + b + c \leq 3.$$
Applying Jensens to $ g(x) = \frac{ 1 } { (a+b+c+1) - x }$, we have
$$ \sum \frac{1}{ b+c+1} = g(a) + g(b) + g(c) \geq 3 g( \frac{ a+b+c} { 3 } ) = 3 \times \frac{ 3 } { 2a+2b+2c + 3 } \geq 1. $$
