By Titu, $\sum_{cyc} \frac{1}{3a+2} \ge \frac{4^2}{3 \cdot 4+8} = \frac{4}{5}$. We have that$\sum_{cyc}\frac{a}{a^3 + 4} \le \sum_{cyc}\frac{a}{3a + 2} = \frac{1}{3} \sum_{cyc}\frac{3a+2-2}{3a + 2} = \frac{4}{3} -\frac{2}{3}\sum_{cyc} \frac{1}{3a+2} \le \frac{4}{3} -\frac{2}{3} \cdot \frac{4}{5} = \frac{4}{5}$

You just need to prove that $$\frac{2(x-1)}{25}+\frac{1}{5}\ge\frac{x}{x^3+4}$$ Which is equivalent to the $$(x^3+4)(2x+3)\ge25x$$Easy $AM \ge GM$ Then add all of them

bump What about the minimum of this?

Since for $x\ge 0$ we have $\frac{1}{x^3+4}\ge -\frac{3}{289}x+\frac{65}{1156}$ hence $$\sum\frac{a}{a^3+4}\ge -\frac{3}{289}\sum a^2+\frac{65}{1156}\sum a\ge -\frac{3}{289}\left(\sum a\right)^2+\frac{65}{1156}\sum a=\frac{1}{17}$$attained for permutations of $(4,0,0,0)$. EDIT: Oh, I have just noticed that $a,b,c,d>0$ in the main problem. In that case either we consider $a,b,c,d\ge 0$ or we are talking about the infimum here.

Observe that $$\frac{a}{a^3+4} \leq \frac{1}{5}+\frac{2}{25}(a-1)$$Factors to $$(a-1)^2(2a^2+7a+12)\geq 0$$Which is obviously true. Summing through $a,b,c,d$ and taking into account that $a+b+c+d=4$ we get the desired.

The following is harder. If $a,b,c,d$ are nonnegative real numbers such that $a+b+c+d=4$, then $$\sum_{cyc}\frac{a}{3a^3 + 2} \le \frac{4}{5}.$$

csav10 wrote: The following is harder. If $a,b,c,d$ are nonnegative real numbers such that $a+b+c+d=4$, then $$\sum_{cyc}\frac{a}{3a^3 + 2} \le \frac{4}{5}.$$ The Vasc's LCF Theorem helps here. Finally, we need to prove that $$42-92a+26a^2+54a^3-27a^4\geq0$$for $0\leq a\leq\frac{4}{3}$ or $$\frac{9}{25}(4-3a)(4-5a)^2(a+1)+\frac{1}{25}(45a^3+542a^2-1004a+474)\geq0,$$which is obvious.

I have used also LCF-Theorem for $f(x)=\frac{-x}{3x^2+2}$ defined on : $I=[0,4]\ $, which is convex on $I_{\le 1}\ $. I shown that $h(x,y)\ge 0$, where $$h(x,y)=\frac{g(x)-g(y)}{x-y},\ \ \ \ g(u)=\frac{f(u)-f(1)}{u-1}.$$We need to show that $x+3y=4$ involves $$2(x^2+xy+y^2)+2(x+y)+2\ge 3x^2y^2+3xy(x+y).$$From $4=x+3y\ge 2\sqrt{3xy}$, it follows that $3xy\le 4$. Thus, it is enough to show that $$2(x^2+xy+y^2)+2(x+y)+2\ge 4xy+4(x+y),$$that is $(y-1)^2\ge 0.$

denery wrote: aplying jensen we get that equality You are welcome to show us your solution.

By applying AM-GM twice the problem will be solved