<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] 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!
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