a,b,c are three positive numbers and abc² has greatest value $\frac{1}{64}$, then

$\begin{array}{l}\frac{\mathrm{a}+\mathrm{b}+\frac{\mathrm{c}}{2}+\frac{\mathrm{c}}{2}}{4}\ge \sqrt[4]{\mathrm{a}\cdot \mathrm{b}\cdot \frac{\mathrm{c}}{2}\cdot \frac{\mathrm{c}}{2}}\text{ i.e. }\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{4}\ge \sqrt[4]{\frac{{\mathrm{abc}}^2}{4}}\\ \therefore \frac{(\mathrm{a}+\mathrm{b}+\mathrm{c}{)}^4}{4^4}=\frac{{\mathrm{abc}}^2}{4}\Rightarrow {\mathrm{abc}}^2\le \frac{1}{64}(\mathrm{a}+\mathrm{b}+\mathrm{c}{)}^4\end{array}$
$\therefore$the greatest value of abc²,
we have $\mathrm{a}=\mathrm{b}=\frac{\mathrm{c}}{2}\therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=1$