Let $a,b,c>0$ and $abc\leq 1$ prove that:
$$3/7\leq {\frac {2-a}{3\,{a}^{3}+4}}+{\frac {2-b}{3\,{b}^{3}+4}}+{ \frac {2-c}{3\,{c}^{3}+4}} $$
I was trying by separating $ (\sum \frac{2}{3a^3+4})-(\sum \frac{a}{3a^3+4})$
Now I need to find the minimum value of $3a^3+4$. I do that by AM-GM as $3a^3+3+1 \ge 3^{\frac{5}{3}}.a$. But that is not doing any good. Do you have better methods?
I didn't find an elegant way to do this, but tried with some brute force. This proof can be checked easily by a calculator. But checking it by hand seems to be too heavy work.
Set $$f(x)=\frac{2-x}{3x^3+4}$$ and $g(x)=f(e^x)$. We get $$f'(x)=-\frac{(3x^3+4)+9x^2(2-x)}{(3x^3+4)^2}=\frac{6x^2(x-3)-4}{(3x^3+4)^2}$$ $$g''(x)=-\frac{2e^x(18e^{6x}-81e^{5x}-78e^{3x}+108e^{2x}+8)}{(3e^{3x}+4)^3}$$ It's not hard to observe that the polynomial $P(x)=18x^6-81x^5-78x^3+108x^2+8$ has exactly two real roots $x_1\approx 0.901789,x_2\approx 4.64096$. We proceed to the following
Proposition 1. $f(x)$ is monotone decreasing on $[0,3]$, and $f(x)>-1/81$ for $x\ge 0$.
Proof. It's easy to see $f'(x)$ has a unique root on $x>0$, which we call $x_0$. It's obvious that $x_0>3$, and since $(3x_0^3+4)+9x_0^2(2-x_0)=0$, we have $$f(x)\ge f(x_0)=\frac{2-x_0}{3x_0^3+4}=-\frac{1}{9x_0^2}>-\frac{1}{81}.$$
Corollary 1. If $0<x\leq 1/6$, $y,z>0$, then $f(x)+f(y)+f(z)>3/7$.
Proof. By Proposition 1 we have $$f(x)+f(y)+f(z)-\frac 3 7\ge f(\frac 1 6)-\frac 1 {81}-\frac 1 {81}-\frac 3 7=\frac{571}{163863}>0.$$
The following two propositions follow from the convexity of $g(x)$.
Proposition 2. For $x,y\in(0,x_1]$ we have $$f(x)+f(y)\ge f(x_1)+f(\frac{xy}{x_1})$$ Proposition 3. For $x,y\in[x_1,x_2]$ we have $$f(x)+f(y)\ge 2f(\sqrt{xy})$$
Now return to the original problem, we claim that
Claim. If $\max(a,b,c)>4.6$, then $f(a)+f(b)+f(c)>3/7$.
Proof. WLOG we may assume $a\le b\le c$. If $c>4.6$ and $b\ge 6/4.6=30/23$, then $a\le 1/bc<1/6$, hence we're done by Corollary 1. Otherwise $b<30/23$. We consider two cases.
Case 1. $b\ge x_1$. Then $a\le 1/bc\le 1/(0.9\times 4.64)$. By Proposition 1 we have $$f(a)+f(b)+f(c)>f(\frac 1{0.9\times 4.64})+f(\frac{30}{23})-\frac 1{81}>\frac 3 7+0.06>\frac 3 7.$$ Case 2. $b<x_1$. Then $a\le b<x_1$. By Proposition 2 we have $$f(a)+f(b)+f(c)\ge f(x_1)+f(\frac{ab}{x_1})+f(c)$$ Now it's reduced to $b=x_1$, which has already been solved in Case 1.This concludes the proof.
Now it suffices to consider the case where $\min(a,b,c)\le 4.6<x_2$, which can be divided into four subcases:
Case 1. $a\le b\le x_1\le c<x_2$. By Proposition 2 we have $$f(a)+f(b)+f(c)\ge f(x_1)+f(\frac{ab}{x_1})+f(c)$$ Now it's reduced to $b=x_1$, which is to be solved in the following Case 2.
Case 2. $a\le x_1\le b\le c<x_2$. Apply Proposition 1, 3 to reduce the problem to proving a polynomial inequality $2f(1/\sqrt{a})+f(a)\ge 3/7$ with the only variable $\sqrt{a}$. This one is easy to deal with. In fact it's equivalent to
$$\frac{(x-1)^2(2x^2+3)(24x^5+27x^4-6x^3-21x^2+4x+2)}{7(4x^3+3)(3x^6+4)}\ge 0$$
where $x=\sqrt{a}$.
Case 3. $a\le b\le c<x_1$. In this case $f(a)+f(b)+f(c)>3f(1)=3/7$.
Case 4. $x_1\le a\le b\le c<x_2$. Apply Proposition 3 (or Jensen's inequality directly) to get the result.
