请听题!

$n\in N_+,$对$\forall x_1,x_2,\cdots,x_n\in R,$有$\sum\limits_{i=1}^n(\sum\limits_{j=1}^ix_j)^2\leq k\cdot\sum\limits_{i=1}^{n}x_i^2,$若$f(n)$为$k$的最小值
求$f(n):$
令 $y_i=\sum\limits_{k=1}^ix_k$
则 原不等式 $\Leftrightarrow \sum\limits_{i=1}^ny_i^2\leq f(n)(y_1^2+(y_2-y_1)^2+\cdots+(y_n-y_{n-1})^2)$
$\Leftrightarrow y_1y_2+y_2y_3+\cdots+y_{n-1}y_n\leq (1-\frac{1}{2f(n)})(y_1^2+y_2^2+\cdots+y_{n-1}^2)+(\frac{1}{2}-\frac{1}{2f(n)})y_n^2$
显然 $f(n)\geq 1$ 不妨设 $cos\alpha=1-\frac{1}{2f(n)}$
原不等式 $\Leftrightarrow y_1y_2+y_2y_3+\cdots+y_{n-1}y_n\leq cos\alpha(y_1^2+y_2^2+\cdots+y_{n-1}^2)+(cos\alpha-\frac{1}{2})y_n^2$
对左边每一项使用均值不等式配系数:
$y_1y_2\leq cos\alpha y_1^2+\frac{1}{4cos\alpha}y_2^2,y_2y_3\leq(cos\alpha-\frac{1}{4cos\alpha})y_2^2+\frac{1}{4(cos\alpha-\frac{1}{4cos\alpha})}y_3^2$
对于 $cos\alpha-\frac{1}{4cos\alpha}$ 有 $cos\alpha-\frac{1}{4cos\alpha}=\frac{4cos^2\alpha-1}{4cos\alpha}=\frac{4-4sin^2\alpha-1}{4cos\alpha}=\frac{3sin\alpha-4sin^3\alpha}{4sin\alpha cos\alpha}=\frac{sin3\alpha}{2sin2\alpha}$
$\frac{1}{4(cos\alpha-\frac{1}{4cos\alpha})}=\frac{sin2\alpha}{2sin3\alpha},y_3y_4\leq(cos\alpha-\frac{sin2\alpha}{2sin3\alpha})y_3^2+\frac{1}{4(cos\alpha-\frac{sin2\alpha}{2sin3\alpha})}y_4^2$
一般地 $cos\alpha-\frac{sink\alpha}{2sin(k+1)\alpha}=\frac{2cos\alpha sin(k+1)\alpha-sink\alpha}{2sin(k+1)\alpha}=\frac{sin(k+2)\alpha+sink\alpha-sink\alpha}{2sin(k+1)\alpha}=\frac{sin(k+2)\alpha}{2sin(k+1)\alpha}$
所以 $y_1y_2\leq\frac{1}{2}(\frac{sin2\alpha}{sin\alpha}y_1^2+\frac{sin\alpha}{sin2\alpha}y_2^2),y_2y_3\leq\frac{1}{2}(\frac{sin3\alpha}{sin2\alpha}y_2^2+\frac{sin2\alpha}{sin3\alpha}y_3^2),\cdots,y_{n-1}y_n\leq\frac{1}{2}(\frac{sinn\alpha}{sin(n-1)\alpha}y_{n-1}^2+\frac{sin(n-1)\alpha}{sinn\alpha}y_n^2)$
此时前 $n-1$ 项的系数完整,考虑第 $n$项的系数
则应当有 $\frac{1}{2}\frac{sin(n-1)\alpha}{sinn\alpha}=cos\alpha-\frac{1}{2}\Leftrightarrow sin(n-1)\alpha+sinn\alpha=2cos\alpha sinn\alpha\Leftrightarrow sin(n-1)\alpha+sinn\alpha=sin(n+1)\alpha+sin(n-1)\alpha\Leftrightarrow sin(n+1)\alpha=sinn\alpha\Leftrightarrow 2cos\frac{2n+1}{2}\alpha sin\frac{\alpha}{2}=0$
所以 $\alpha=\frac{\pi}{2n+1},f(n)=\frac{1}{2-2cos\frac{\pi}{2n+1}}$