<s>[INST] Prove that $\frac{a}{b^2+5}+ \frac{b}{c^2+5} + \frac{c}{a^2+5} \le \frac 12$ Let $a,b,c>0$ and $a^3+b^3+c^3=3$. Prove that $$\dfrac{a}{b^2+5}+ \dfrac{b}{c^2+5} + \dfrac{c}{a^2+5} \le \dfrac 12$$
I have an ugly solution for this solution.
 [/INST] First ,we can get rid of denominators, by noticing that 
$\frac{1}{x^2+5} \leq \frac{4-x}{18}$ for any $x\in [0,2]$
(indeed, we have
$$
(4-x)(x^2+5)-18=((x-1)^2)(2-x) \geq 0 \tag{1}
$$
So it will suffice to show that the number
$$
f(a,b,c)=a(4-b)+b(4-c)+c(4-a)  \tag{2}
$$
is $\leq 9$. Note that $f(a,b,c)=4(a+b+c)-(ab+ac+bc)$ is fully symmetric
in $a,b,c$. 
We need a “lower-dimensional” result :
 Lemma.  Let $\alpha,\beta$ be two positive numbers such that $\alpha^3+\beta^3=2$.
Then $\alpha(3-\beta)+\beta(3-\alpha) \leq 4$, with equality iff $\alpha=\beta=1$.
 Proof of lemma  The proposed inequality is equivalent to
$\beta(3-2\alpha) \leq 4-3\alpha$, or $(2-\alpha^3)(3-2\alpha)^3 \leq (4-3\alpha)^3$. Now
$$
(4-3\alpha)^3-(2-\alpha^3)(3-2\alpha)^3=(\alpha-1)^4 \bigg(\frac{42}{25}+
\big(\frac{13}{10}-\alpha\big)\big(8\alpha+\frac{32}{5}\big)\bigg),
$$
which concludes the proof of the lemma.
Now let $a,b,c$ be positive numbers with $a^3+b^3+c^3=1$. Set $t=\big(\frac{b^3+c^3}{2}\big)^{\frac{1}{3}}$, so that $b^3+c^3=2t^3$. Then the numbers
$\alpha=\frac{b}{t}$ and $\beta=\frac{c}{t}$ satisfy the hypotheses of the lemma ; we deduce
$$
b(3t-c)+c(3t-b) \leq 4t^2, \ \text{with equality iff } \ b=c=t \tag{3}
$$ 
This means that
$$
f(a,b,c) \leq f(a,t,t), \ \text{with equality iff } \ b=c=t \tag{4}
$$
The set $K=\lbrace (a,b,c) \in [0,+\infty[^3 | a^3+b^3+c^3=1\rbrace$ is compact, so
the continuous map $f$ attains its maximum on $K$ at some point $(a_0,b_0,c_0)$. Then (4)
shows that we must have $b_0=c_0$, and by symmetry $a_0=b_0=c_0=1$, qed.
</s>