By the AM-GM inequality, we have
$$(a + b + 4c)^2 \geq (2\sqrt{(a+b)(4c)})^2 = 16(a+b)c$$
and
$$\frac{100abc}{a+b+c} \leq \frac{100c}{a+b+c} \left(\frac{a+b}{2}\right)^2 = \frac{25(a+b)^2c}{a+b+c}.$$
Therefore, it suffices to prove
$$d^2 + 16cd \geq \frac{25cd^2}{c+d}$$
where $d = a + b$. Indeed,
$$\begin{align*} d^2 + 16cd &\ge \frac{25cd^2}{c+d} \\ \Leftrightarrow \quad d(16c+d)(c+d) &\ge 25cd^2 \\ \Leftrightarrow \quad d(16c^2 - 8cd + d^2) &\ge 0 \\ \Leftrightarrow \quad d(4c-d)^2 &\ge 0. \end{align*}$$
This is obviously true. Equality holds when $a = b = 2c$.