If $a+b+c \ge 1$ and $a,b,c>0$, prove that $\frac{1}{2a+s}+\frac{1}{2b+s}+\frac{1}{2c+s} \ge \frac{1}{s}$, where $s=ab+bc+ca$. What I know is that $$s \le \frac{(a+b+c)^2}{3} \ge \frac{1}{3}$$But as you can see the sign is pointing to different sides. So I can't see how this could be helpful. Just a small observation. I don't have any other ideas and I'd like to get some help. Thanks.

first we prove $a+b+c=1$ case:
$\dfrac{1}{2a+s}+\dfrac{1}{2b+s}+\dfrac{1}{2c+s} \ge \dfrac{1}{s} \iff s^3+s^2-4abc\ge 0$
let $x=3a,y=3b,z=3c,3u=x+y+z=3,3v^2=xy+yz+xz=9s,w^3=xyz=27abc$
$s^3+s^2-4abc\ge 0 \iff v^6+3v^4-4w^3 \ge 0 \iff v^6+3v^4\ge 4w^3$
accoding to $uvw$ method, $w^3\le 3v^2-2+2\sqrt{(1-v^2)^3}$
$\iff v^6+3v^4 \ge 12v^2-8+8\sqrt{(1-v^2)^3} \iff v^6+3v^4 -12v^2+8\ge 8\sqrt{(1-v^2)^3} \iff (v^6+3v^4 -12v^2+8)^2 \ge 64(1-v^2)^3 \iff v^{12}+6v^{10}-15v^8+8v^6 \ge 0 \iff v^6+6v^4-15v^2+8 \ge 0 \iff (v-1)^2(v^2+8)(v+1)^2\ge 0$
when $v=1$ the "="hold $\implies x=y=z=1 \implies a=b=c=\dfrac{1}{3}$
for $a'+b'+c'>1$, let$ a'+b'+c'=k(a+b+c) \to k > 1, ,a'=ka,b'=kb,c'=kc,s'=k^2s$
$\dfrac{1}{2a'+s'}+\dfrac{1}{2b'+s'}+\dfrac{1}{2c'+s'} \ge \dfrac{1}{s'} \iff \dfrac{1}{2\frac{a}{k}+s}+\dfrac{1}{2\frac{b}{k}+s}+\dfrac{1}{2\frac{c}{k}+s} \ge \dfrac{1}{s}$
$\frac{a}{k} < a \implies \dfrac{1}{2\frac{a}{k}+s}>\dfrac{1}{2a+s} $ so the inequlity hold.
QED.
