Proving an inequality for positive numbers $a, b, c$ 
Let be $a,b,c$ positive numbers such that $a+b+c=3$. Prove that 
  $$\frac{b+c+bc}{a^2+b^3+c^4}+\frac{c+a+ca}{b^2+c^3+a^4}+\frac{a+b+ab}{c^2+a^3+b^4} \le 3$$

 A: Using Cauchy Schwarz inequality:
$(a^2+b^3+c^4)(a^2+b+1) \ge (a^2+b^2+c^2)^2$. Do this for each denominator, then it suffices to show that 
$$\sum (a^2+b+1)(b+c+bc) \leq 3(a^2+b^2+c^2)^2$$
where the sum is cyclic. Expand the left hand side, we get
$$\begin{eqnarray}
&\sum a^2b + b^2 + b + a^2c + bc + c + a^2bc + b^2c + bc \leq 3(a^2+b^2+c^2)^2 \\
\Leftrightarrow & 2 \sum a^2b + \sum a^2 c + \sum b^2 + \sum b + \sum c + 2\sum bc + \sum a^2bc \leq 3(a^2+b^2+c^2)^2 \\
\Leftrightarrow & \sum (a^2b+a^2c) + \sum a^2b + \sum a^2 + 3 + 3 + 2\sum ab + 3abc \leq 3(a^2+b^2+c^2)^2 \cdots (*)
\end{eqnarray}$$
using $a+b+c = 3$. Now note the identity 
$$\sum(a^2b+a^2c) +3abc = a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+3abc = (ab+bc+ca)(a+b+c)$$
which is equal to $3(ab+bc+ca)$ for this problem, so
$$(*)\Leftrightarrow \sum a^2 + 5 \sum ab + 6 + \sum a^2b \leq 3(a^2+b^2+c^2)^2$$
to be proved now. 
Note the following inequalities:
$$\begin{eqnarray}(1) & \hspace{5mm}(a^2+b^2+c^2)^2 +3 \ge 4 \sum a^2b \\
(2)& \hspace{5mm}(a^2+b^2+c^2) \ge 3 \\
(3)& \hspace{5mm}(a^2+b^2+c^2) \ge ab+bc+ca
\end{eqnarray}$$
Here (1) follows from applying AM-GM to $a^4+a^2b+a^2b+1 \ge 4a^2b$ and sum up cyclicly. The other two are routine applications of Cauchy-Schwarz/AM-GM.
These implies
$$\begin{eqnarray}(1') &\hspace{5mm} \frac{(a^2+b^2+c^2)^2}{3} \ge \frac{1}{4} (a^2+b^2+c^2)^2 + \frac{3}{4} \ge \sum a^2b \\
(2')&\hspace{5mm} \frac{2}{3}(a^2+b^2+c^2)^2 \ge 6 \\
(3')&\hspace{5mm} \frac{5}{3}(a^2+b^2+c^2)^2 \ge 5(a^2+b^2+c^2) \ge 5\sum ab \\
(4')&\hspace{5mm} \frac{1}{3} (a^2+b^2+c^2)^2 \ge \sum a^2 \end{eqnarray}$$
The inequality then follows from $(1') + (2') + (3') + (4')$.
A: As a habit I write x,y,z for a,b,c. Letting $z = 3 - (x+y)$ and plotting the function in Mathematica reveals a local maximum on $[\{0,3\},\{0,3\}]$ which does appear to be very close to $x= y = 1,$ and the maximum attained is of course 3. 
The partial derivative of the l.h.s. of the OP with respect to y letting x = 1 is zero at y = 1. The partial derivative of the l.h.s. with respect to x at y = 1 is zero at x = 1. 
On the domain of interest, it does then appear that $a = b = c= 1$ give a maximum of 3 and the objective function is less than three for $(x,y)\neq (1,1)$ and so the inequality is true. 
A: By Holder
$$\sum_{cyc}\frac{a+b+ab}{c^2+a^3+b^4}=\sum_{cyc}\frac{9(c^2+a+1)(a+b+ab)}{(c^2+a+1)(c^2+a^3+b^4)(1+1+1)(1+1+1)}\leq$$
$$\leq\frac{9\sum\limits_{cyc}(c^2+a+1)(a+b+ab)}{(a+b+c)^4}=\frac{1}{9}\sum_{cyc}(a^2b+a^2c+a^2bc+a^2+ab+a^2b+2+ab)=$$
$$=\frac{1}{9}\sum_{cyc}(2a^2b+a^2c+a^2bc+a^2+2ab+2)=\frac{1}{9}\sum_{cyc}(2a^2b+a^2c+abc+5)=$$
$$=\frac{2(a^2b+b^2c+c^2a+abc)+(a^2c+b^2a+c^2b+abc)+15}{9}\leq\frac{2\cdot4+4+15}{9}=3.$$
We used the following lemma.

Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3$. Prove that:
  $$a^2b+b^2c+c^2a+abc\leq4$$

A proof of the lemma.
Let $\{a,b,c\}=\{x,y,z\}$ such that $x\geq y\geq z$.
Hence, by Rearrangement and AM-GM we obtain:
$$a^2b+b^2c+c^2a+abc=a\cdot ab+b\cdot bc+c\cdot ca+xyz\leq x\cdot xy+y\cdot xz+z\cdot yz+xyz=$$
$$=y(x+z)^2=4y\left(\frac{x+z}{2}\right)^2\leq4\left(\frac{y+\frac{x+z}{2}+\frac{x+z}{2}}{3}\right)^3=4$$
Done!
