If $a,b,c\ge 0:ab+bc+ca>0$ which sum is equal to $2$. Prove $$abc+\frac{1}{ab+bc+ca}\ge 1$$ I've tried to use Schur without success. I am waiting a proof without $uvw$
Let $0\le r\le \dfrac{8}{27}$ and $0<q\le \dfrac{4}{3}.$
Now, $r\ge \dfrac{p(4q-p^2)}{9}=\dfrac{8(q-1)}{9},$ and we need prove$$\frac{1}{q}+\dfrac{8(q-1)}{9}\ge 1 \iff \frac{(8q-9)(q-1)}{q}\ge 0,$$which is wrong when $\dfrac{9}{8}> q > 1.$
I also saw that $uvw$ helps well.
Actually, by the method and for fixed $u,v$ the min of $LHS$ occurs when either $abc=0$ or two of $a,b,c$ are equal.
WLOG $abc=0\implies a=0$ and it is $\frac1{bc}\ge1$ where $b+c=2$. Easy. Equality at $a=0, b=c=1 \quad\blacksquare$
WLOG let $b=c$ and it is $ab^2+\frac1{b^2+2ab} \ge1$ where $a+2b=2$
Eliminate $a$ and you get $\frac{6 b^5 - 14 b^4 + 8 b^3 + 1}{b(4-3b)} \ge1$ where $0\le b \le1 $
$\frac{6 b^5 - 14 b^4 + 8 b^3 + 1}{b(4-3b)}-1=\frac{ (b-1)^2(6b^3 - 2b^2 - 2b + 1)}{b (4-3b)} \ge0$ at $b=c=1, a=0\quad\blacksquare$
$\left[6b^3-2b^2-2b+1 = (2+6b)\left(b-\frac12\right)^2+2\left(b-\frac38\right)^2+\frac7{32}>0\right]$
Plus cyclic variations and we get result.
Also, we can rewrite the inequality $$\frac{(a+b+c)^5}{16(ab+bc+ca)}+2abc\ge \frac{\left( a+b+c \right) ^3}{4},\forall a,b,c\ge 0: ab+bc+ca>0$$There is a equivalent SOS expression $$\sum_{\mathrm{cyc}}{\left[ a^2\left( a+2b+2c \right) \left( b+c-a \right) ^2+a\left( b-c \right) ^2\left( b+c-a \right) ^2 \right]}\ge 0.$$
By Schur of fourth degree, $r\ge \dfrac{(4q-p^2)(p^2-q)}{6p}=\dfrac{5q-q^2-4}{3}.$
Now, we only need to prove$$\frac{1}{q}+\frac{5q-q^2-4}{3}\ge 1 \iff \frac{(3-q)(q-1)^2}{q}\ge 0.$$Hence, we get Q.E.D
Equality occurs when $(a,b,c)=(0,1,1)$ and cylic permutations. About information of Schur inequality, see here.