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!

Because by AM-GM we have:
$$\frac{a^2}{a^3+5}\leq\frac{a}{4}$$
$$a^3+2\cdot2.5\geq3\sqrt[3]{a^3\cdot2.5^2}>4a.$$
