Lời giải:
$M=c^2(\frac{1}{a^2}+\frac{1}{b^2})+\frac{a^2+b^2}{c^2}+2017$
$\geq \frac{4c^2}{a^2+b^2}+\frac{a^2+b^2}{c^2}+2017$ (theo BĐT Cauchy-Schwarz)
$=\frac{3c^2}{a^2+b^2}+(\frac{c^2}{a^2+b^2}+\frac{a^2+b^2}{c^2})+2017$
$\geq \frac{3(a^2+b^2)}{a^2+b^2}+2\sqrt{\frac{c^2}{a^2+b^2}.\frac{a^2+b^2}{c^2}}+2017=3+2+2017=2022$ (theo BĐT AM-GM)
Vậy $M_{\min}=2022$