设数列\(\{a_{n}\}(n=1,2,3…)\)的前\(n\)项和\(S_{n}\)满足\(S_{n}=2a_{n}-a_{1}\)，且\(a_{1}\)，\(a_{2}+1\)，\(a_{3}\)成等差数列，数列\(\{b_{n}\}\)满足\(a_{1}\)，\(a_{2}\)，\(a_{3}…{{a}_{n}}={{(\sqrt{2})}^{bn}}(n\in {{N}^{*}})\)．
\((1)\)求\(a_{n}\)与\(b_{n}\)；
\((2)\)设\({{c}_{n}}=\dfrac{1}{{{a}_{n}}}-\dfrac{1}{{{b}_{n}}}(n\in {{N}^{*}})\)，记数列\(\{c_{n})\)的前\(n\)项和为\(T_{n}.\)求证：对任意\(n∈N^{*}\)，均有\(T_{n} > 0\)．
\((3)\)设\({{d}_{n}}={{b}_{n}}-n(n\in {{N}^{*}})\)，\(f(n)=\dfrac{1}{\sqrt{n+{{d}_{1}}}}+\dfrac{1}{\sqrt{n+{{d}_{2}}}}+\cdots +\dfrac{1}{\sqrt{n+{{d}_{n}}}}(n\in {{N}^{*}},n\geqslant 2)\)，求\(f(n)\)的最小值．