Let $f(a, b, c) = \frac{a^4 + 5b^4}{a(a + 2b)} + \frac{b^4 + 5c^4}{b(b + 2c)} + \frac{c^4 + 5a^4}{c(c + 2a)}$. Since $a + b + c = 1$ we will prove that $f(a, b, c) \ge a^2 + b^2 + c^2 + ab + bc + ac$.

By Cauchy-Schwarz inequality for positive $x_1, \dots, x_n$
$$(x_1 + \dots + x_n) \left( \frac{a_1^2}{x_1} + \dots + \frac{a_n^2}{x_n} \right) \ge (a_1 + \dots + a_n)^2 \quad (1)$$
By applying (1) we get
$$\frac{a^4}{a(a + 2b)} + \frac{b^4}{b(b + 2c)} + \frac{c^4}{c(c + 2a)} \ge \frac{(a^2 + b^2 + c^2)^2}{(a + b + c)^2} = (a^2 + b^2 + c^2)^2 \quad (2)$$
$$\frac{b^4}{a(a + 2b)} + \frac{c^4}{b(b + 2c)} + \frac{a^4}{c(c + 2a)} \ge \frac{(a^2 + b^2 + c^2)^2}{(a + b + c)^2} = (a^2 + b^2 + c^2)^2 \quad (3)$$
Now (2) + 5 (3) gives $f(a, b, c) \ge 6(a^2 + b^2 + c^2)^2$. Thus, the proof of
$$6(a^2 + b^2 + c^2)^2 \ge a^2 + b^2 + c^2 + ab + bc + ac \quad (4)$$
will complete the solution. Now the sum of $\frac{a^2 + b^2}{2} \ge ab$, $\frac{b^2 + c^2}{2} \ge bc$, $\frac{c^2 + a^2}{2} \ge ca$ gives $a^2 + b^2 + c^2 \ge ab + bc + ca$. Therefore, (4) will follow from $6(a^2 + b^2 + c^2)^2 \ge 2(a^2 + b^2 + c^2)$ or $3(a^2 + b^2 + c^2)^2 \ge 1$. But since $a+b+c=1$ the last inequality is a quadratic-arithmetic means inequality. Done.