%2007/04/22 06:10
%Copyright 2007 WATANABE,Masayuki
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\begin{document}
{\large 京都大学　1975年　入学試験　理系数学　問題2}\\
\\
{\large 問題}\\
\\
\begin{enumerate}
 \item　$x\geqq 0$ のとき $\sqrt{x}\leqq\dfrac{1}{2}(x+1)$ を示せ．
 \item　$\dint_0^k \sqrt{x}\Bigl(1-\dfrac{x}{k}\Bigr)^k\,dx<1$ を示せ（$k$は自然数）．
\end{enumerate}
\ \\
{\large 解答}\\
\\
1.\\
相加平均$\geqq$相乗平均\\
より\\
$\dfrac{a+b}{2}\geqq \sqrt{ab}$\\
$x\geqq 0$なので\\
$a=1,b=x$とおくと\\
$\dfrac{1+x}{2}\geqq \sqrt{x}$\\
となり、\\
題意は成立する。\\
2.\\
$S_k=\dint_0^k \sqrt{x}\Bigl(1-\dfrac{x}{k}\Bigr)^k\,dx<1$\\
$S_k\leqq \dint_0^k \dfrac{x+1}{2}\Bigl(1-\dfrac{x}{k}\Bigr)^k\,dx$\\
$T_k=\dint_0^k (x+1)(k-x)^k\,dx$とおく\\

$\dint f(x)g'(x)\,dx=f(x)g(x)-\dint f'(x)g(x)\,dx$\\
$f(x)=(x+1)$\\
$g'(x)=(k-x)^k$\\
$f'(x)=1$;
$g(x)=-\dfrac{(k-x)^{k+1}}{k+1}$\\
\begin{eqnarray*}
T_k&=&\dint_0^k (x+1)(k-x)^k\,dx\\
&=&\left[-\dfrac{(x+1)(k-x)^{k+1}}{k+1}\right]_0^k-\dint_0^k -\dfrac{(k-x)^{k+1}}{k+1}\,dx\\
&=&-\dfrac{(k+1)(k-k)^{k+1}}{k+1}+\dfrac{(0+1)(k-0)^{k+1}}{k+1}+\dint_0^k \dfrac{(k-x)^{k+1}}{k+1}\,dx\\
&=&\dfrac{k^{k+1}}{k+1}+\dint_0^k \dfrac{(k-x)^{k+1}}{k+1}\,dx\\
&=&\dfrac{k^{k+1}}{k+1}+\left[-\dfrac{(k-x)^{k+2}}{(k+1)(k+2)}\right]_0^k\\
&=&\dfrac{k^{k+1}}{k+1}+-\dfrac{(k-k)^{k+2}}{(k+1)(k+2)}+\dfrac{(k-0)^{k+2}}{(k+1)(k+2)}\\
&=&\dfrac{k^{k+1}}{k+1}+\dfrac{k^{k+2}}{(k+1)(k+2)}\\
&=&\dfrac{k^{k+1}}{k+1}\left(1+\dfrac{k}{(k+2)}\right)
\end{eqnarray*}
\begin{eqnarray*}
\dfrac{1}{2k^k}T_k&=&\dfrac{1}{2k^k}\dfrac{k^{k+1}}{k+1}\left(1+\dfrac{k}{(k+2)}\right)\\
&=&\dfrac{k}{2(k+1)}\left(1+\dfrac{k}{(k+2)}\right)\\
&=&\dfrac{k(2k+2)}{2(k+1)(k+2)}\\
&=&\dfrac{k(k+1)}{(k+1)(k+2)}\\
&=&\dfrac{k}{k+2}
\end{eqnarray*}
以上より
$$S_k<\dfrac{k}{k+2}<1$$
よって題意は成立する。\\
\end{document}

