20.观察下面的解答过程:已知正实数a,b满足a+b=1,求$\sqrt{2a+1}$+$\sqrt{2b+1}$的最大值.
解:∵$\sqrt{2a+1}$•$\sqrt{2}$≤$\frac{(\sqrt{2a+1})^{2}+{\sqrt{2}}^{2}}{2}$=a+$\frac{3}{2}$,$\sqrt{2b+1}$•$\sqrt{2}$≤$\frac{{\sqrt{2b+1}}^{2}{+\sqrt{2}}^{2}}{2}$=b+$\frac{3}{2}$,
相加得$\sqrt{2a+1}$•$\sqrt{2}$+$\sqrt{2b+1}$•$\sqrt{2}$=$\sqrt{2}$•($\sqrt{2a+1}$+$\sqrt{2b+1}$)≤a+b+3=4,
∴$\sqrt{2a+1}$+$\sqrt{2b+1}$≤2$\sqrt{2}$,等号在a=b=$\frac{1}{2}$时取得,即$\sqrt{2a+1}$+$\sqrt{2b+1}$的最大值为2$\sqrt{2}$.
请类比以上解题法,使用综合法证明下题:
已知正实数x,y,z满足x+y+z=3,求$\sqrt{2x+1}$+$\sqrt{2y+1}$+$\sqrt{2z+1}$的最大值.