[tex]\it a=3\sqrt2<3\sqrt4=3\cdot2=6<20[/tex]
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[tex]\it b=\dfrac{2}{\sqrt2}+\dfrac{1}{\sqrt3}+\sqrt2+\dfrac{3}{\sqrt3}=\dfrac{2}{\sqrt2}+\sqrt2+\dfrac{4}{\sqrt3}\\ \\ \\ a\cdot b=3\sqrt2\Big(\dfrac{2}{\sqrt2}+\sqrt2+\dfrac{4}{\sqrt3}\Big)=3\cdot2+3\cdot2+\dfrac{^{\sqrt3)}12\sqrt2}{\sqrt3}=\\ \\ \\ =12+\dfrac{12\sqrt6}{3}=12+4\sqrt6[/tex]
