1.
二次函数\(f(x)\)满足\(f(0)=f(1)=0\),且最小值是\(- \dfrac {1}{4}\).
\((1)\)求\(f(x)\)的解析式;
\((2)\)设常数\(t∈(\;0\;,\; \dfrac {1}{2}\;)\),求直线:\(y=t^{2}-t\)与\(f(x)\)的图象以及\(y\)轴所围成封闭图形的面积是\(S(t)\);
\((3)\)已知\(m\geqslant 0\),\(n\geqslant 0\),求证:\( \dfrac {1}{2}(\;m+n\;)^{2}+ \dfrac {1}{4}(\;m+n\;)\geqslant m \sqrt {n}+n \sqrt {m}\).