21. 已知正实数aaa,bbb满足a>ba>ba>b,且ab=12ab=\frac{1}{2}ab=21,则4a2+b2+12a−b\frac{4{a}^{2}+{b}^{2}+1}{2a-b}2a−b4a2+b2+1的最小值为___232\sqrt{3}23___.