解答题

\((1)\)已知\(x+y+z=1\)，求证：\({{x}^{2}}+{{y}^{2}}+{{z}^{2}}\geqslant \dfrac{1}{3}\)．

\((2)\)已知\(a > 0\)，\(\dfrac{1}{b}-\dfrac{1}{a} > 1\)，求证：\(\sqrt{1+a} > \dfrac{1}{\sqrt{1-b}}\)．