It was a Saturday!

$\begin{aligned} 21x^3+5y^2+2016z^2 &=\frac{21x^3}{2}+\frac{21x^3}{2}+\frac{5y^2}{3}+\frac{5y^2}{3}+\frac{5y^2}{3}+672z^2+672z^2+672z^2 \\ &\ge 8\left(\left(\frac{21x^3}{2}\right)^2 \left(\frac{5y^2}{3}\right)^3 (672z^2)^3 \right)^{\frac{1}{8}} \\&= 8\left(2^{13}3^25^37^5\left(xyz\right)^6\right)^{\frac{1}{8}} \end{aligned}$

Since $xyz=\frac{5\sqrt{70}}{\sqrt[3]{3}} = 5^{\frac{3}{2}} 7^{\frac{1}{2}}2^{\frac{1}{2}} 3^{\frac{-1}{3}}$ , then $(xyz)^6 = 5^97^32^33^{-2}$ . Now

$21x^3+5y^2+2016z^2 \ge 8\left(2^{16}5^{12}7^8\right)^{\frac{1}{8}} =1120\sqrt{5}$ The equality holds if and only if $\frac{21x^3}{2}=\frac{5y^2}{3}=672z^2$ , which eventually implies that $x=3^{\frac{-1}{3}} 2^15^{\frac{1}{2}}$ , $y=3^{\frac{1}{2}}7^{\frac{1}{2}} 2^15^{\frac{1}{4}}$ and $z=3^{\frac{-1}{2}} 2^{\frac{-3}{2}}5^{\frac{1}{2}}$ .

So $a=1120$ and $b=5$ and thus $a+b=\boxed{1125}$ .