Nov 162017
 

不是完整的试题解答, 仅仅在关隘的地点聊一聊.

付云皓的题 3 的解

记 \(a=[nq^{\frac13}]\), \(b=[nq^{\frac23}]\), \(c=nq\). 然后

\begin{equation}\Big(c-aq^{\frac23}\Big)^2+\Big(c-bq^{\frac13}\Big)^2+\Big(aq^{\frac23}-bq^{\frac13}\Big)^2=\frac{2(a^3q^2+b^3q+c^3-3abcq)}{aq^{\frac23}+bq^{\frac13}+c}\geqslant\frac2{3c},\end{equation}

最后的不等式是因为 \(a^3q^2+b^3q+c^3\gt 3abcq\), 并且 \(c\geqslant aq^{\frac23}\), \(c\geqslant bq^{\frac13}\).

然后, 因为 \(c-aq^{\frac23}\geqslant0\), \(c- bq^{\frac13}\geqslant0\), 以及 \(aq^{\frac23}-bq^{\frac13}=-\Big((c-aq^{\frac23})-(c-bq^{\frac13})\Big)\), 得到

\begin{equation}\Big(c-aq^{\frac23}\Big)^2+\Big(c-bq^{\frac13}\Big)^2\leqslant\Big(2c-aq^{\frac23}-bq^{\frac13}\Big)^2,\end{equation}

\begin{equation}\Big(aq^{\frac23}-bq^{\frac13}\Big)^2\leqslant\Big(2c-aq^{\frac23}-bq^{\frac13}\Big)^2.\end{equation}

现在, \((1)\) 给出

\begin{equation}2\Big(2c-aq^{\frac23}-bq^{\frac13}\Big)^2\geqslant\frac2{3c},\end{equation}

记得 \(c=nq\), 这也就是

\begin{equation}\Big(q^{\frac13}\cdot \{nq^{\frac23}\}+q^{\frac23}\cdot \{nq^{\frac13}\}\Big)^2\geqslant\frac1{3nq}.\end{equation}

随即我们有

\begin{equation}\Big\{nq^{\frac23}\Big\}+ \Big\{nq^{\frac13}\Big\}\geqslant\frac1{q\sqrt{3qn}}.\end{equation}

 Posted by at 8:39 pm  Tagged with:

 Leave a Reply

(required)

(required)