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

 [/INST] 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 $(1') + (2') + (3') + (4')$.
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