Prove the inequalitiy 
If $a, b, c$ are positive reals such that:
  $\frac a {b + c + 1} + \frac b {a + c + 1} + \frac c {b + a + 1} \le
 1$ then:
$\frac 1 {b + c + 1} + \frac 1 {a + c + 1} + \frac 1 {b + a + 1} \ge
 1$


I tried using inequality of arithmetic and harmonic means, to no avail.
Any help is appreciated.

.WLOG, $a <b <c$, then $\frac{1}{b+c} < \frac{1}{a+c} < \frac{1}{a+b}$
Note that $$\frac{a}{b+c+1} + \frac{b}{a+c+1} + \frac{c}{a+b+1} = (a+b+c+1)\bigg(\frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1}\bigg) - 3$$
Hence, 
$$
(a+b+c+1)\bigg(\frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1}\bigg) \leq 4
$$
Further, by the harmonic mean-arithmetic mean inequality,
$$
\bigg(\frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1}\bigg) \geq \dfrac{9}{2(a+b+c)+3}
$$
Now, let $x=\bigg(\frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1}\bigg) $ and $y = a+b+c$. Then, we have: $2yx+3x \geq 9$ and $4 \geq yx + x$. Multiplying by $2$ and changing the sign, $-2yx-2x \geq 8$. Adding ,we get $x \geq 1$. Hence the result.
