Find Minimum value of this expression: $P=2(a^3+b^3+c^3)+4(ab+bc+ca)+abc$ Let $a,b,c>0$ and satisfying $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le\frac{a+b+c}{2}$. Find Min of this expression? 
$P=2(a^3+b^3+c^3)+4(ab+bc+ca)+abc$
From the condition, I have :$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{9}{a+b+c+3}$
So, I get: $\frac{a+b+c}{2}\ge\frac{9}{a+b+c+3}\Leftrightarrow (a+b+c-3)(a+b+c+6)\ge 0\Leftrightarrow a+b+c\ge 3$
And then, I can't find the way to mutate P with one variable... Please guide me. Thanks

first, we prove when $a+b+c=3$ ,$P$ will get min when $a=b=c=1$
Let $3u=a+b+c=3,3v^2=ab+bc+ac,w^3=abc, \to u=1 \ge v \ge w, w^3\ge3v^2-2-2\sqrt{(1-v^2)^3}$
$(1-v^2)\le 1 \implies \sqrt{(1-v^2)^3} \le (1-v^2) \implies w^3\ge3v^2-2-2(1-v^2) =5v^2-4$, 
$P=54-42v^2+7w^3 \ge 26-7v^2\ge 19$ when $v=1=u \implies a=b=c=1, P_{min}=19 $
if $a'+b'+c'>3,a'+b'+c'=k(a+b+c) \implies k>1,a'=ka,b'=kb,c'=kc$
$P'=2(a'^3+b'^3+c'^3)+4(a'b'+b'c'+c'a')+a'b'c'=k^32(a^3+b^3+c^3)+k^24(ab+bc+ca)+k^3abc>P=2(a^3+b^3+c^3)+4(ab+bc+ca)+abc \ge 19$
QED.
