Prove that if $a,b,c > 0$ and $a + b + c = 1$, we have: $\frac{a^2}{a^3 + 5} + \frac{b^2}{b^3 + 5} + \frac{c^2}{c^3 + 5} \leq \frac{1}{4}$ Prove that if $a,b,c > 0$ such that $a + b + c = 1$, then the following inequality holds:
$$S = \frac{a^2}{a^3 + 5} + \frac{b^2}{b^3 + 5} + \frac{c^2}{c^3 + 5} \leq \frac{1}{4}$$
What I have tried so far is the following:
Firstly I rewrote $S$ as:
$$ S = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 5\left[ \frac{1}{a(a^3 + 5)} + \frac{1}{b(b^3 + 5)} + \frac{1}{c(c^3 + 5)}\right]$$ 
Then, in order to upper bound $S$, I used the inequality: $ \frac{x^2}{u} + \frac{y^2}{v} + \frac{z^2}{w} \geq \frac{(x + y + z)^2}{u + v + w} $ for any $u,v,w > 0$. 
Therefore, I got:
$$ S \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 5 \cdot \frac{(1 + 1 + 1)^2}{a^4 + b^4 + c^4 + 5(a + b + c)} = \frac{45}{a^4 + b^4 + c^4 + 5}$$
Then, given what we want to show about $S$, this would reduce to proving that:
$$ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{185 + a^4 + b^4 + c^4}{4(a^4 + b^4 + c^4 + 5)} $$
At which point I got stuck and I am not sure whether I started the right way. 
I would be grateful for any suggestions. Many thanks!

Based on Michael Rozenberg's idea, we also have:
$$\frac{a^2}{a^3+5}=\frac{a}{6}-\frac{a(1-a)(5-a^2-a)}{a^3+5}\leq \frac{a}{6}$$
so that:
$$\frac{a^2}{a^3+5}+\frac{b^2}{b^3+5}+\frac{c^2}{c^3+5}\leq \frac{1}{6}(a+b+c)=\frac{1}{6}$$
Equality is attained when $(a,b,c)=(1,0,0)$ up to any cyclic permutation. 
We can use the tangent line method to find the minimum:
$$
\begin{aligned}
\frac{a^2}{a^3+5}&=\frac{3}{136}+\frac{2421}{18496}\left(a-\frac{1}{3}\right)+\frac{(3a-1)^2(1995-135a-269a^2)}{18496(a^3+5)}\\
&\geq \frac{3}{136}+\frac{2421}{18496}\left(a-\frac{1}{3}\right)
\end{aligned}
$$
so that the minimum value is $\frac{9}{136}$ when $a=b=c=\frac{1}{3}$. Therefore the entire range of the left hand side is $\left[\frac{9}{136},\frac{1}{6}\right]$.
