math-ai/BlueMO

1{2    "source_file": "./raw_volume-zh/volume5/exercise8.tex",3    "problem_type": "calculation",4    "problem": "问题12. 设 $x_1, x_2, \\cdots, x_n$ 均不小于 0 , 且 $\\sum_{i=1}^n x_i=1$, 求和式\n$$\n\\sum_{1 \\leqslant i<j \\leqslant n} x_i x_j\\left(x_i+x_j\\right)\n$$\n的最大值.",5    "solution": "$\\sum_{1 \\leqslant i<j \\leqslant n} x_i x_j\\left(x_i+x_j\\right)=\\frac{1}{2} \\sum_{1 \\leqslant i<j \\leqslant n} x_i x_j\\left(x_i+x_j\\right)+\\frac{1}{2} \\sum_{1 \\leqslant i<j \\leqslant n} x_j x_i\\left(x_j+\\right.\\left.x_i\\right)=\\frac{1}{2} \\sum_{i \\neq j} x_i x_j\\left(x_i+x_j\\right)=\\frac{1}{2} \\sum_{i \\neq j} x_i^2 x_j+\\frac{1}{2} \\sum_{i \\neq j} x_i x_j^2=\\frac{1}{2} \\sum_{i=1}^n x_i^2\\left(1-x_i\\right)+ \\frac{1}{2} \\sum_{j=1}^n x_j^2\\left(1-x_j\\right)=\\sum_{i=1}^n x_i^2-\\sum_{i=1}^n x_j^3$.\n当 $n=1$ 时, $x_1=1$, 上式右端为 0 .\n当 $n=2$ 时, 上式右端 $=\\left(x_1^2+x_2^2\\right)-\\left(x_1^3+x_2^3\\right)$.\n$x_1 x_2 \\leqslant \\frac{1}{4}\\left(x_1+x_2\\right)^2=\\frac{1}{4}$, 当 $x_1=x_2=\\frac{1}{2}$ 时, 等号成立.\n当 $n \\geqslant 3$ 时,不妨设 $x_1 \\geqslant x_2 \\geqslant \\cdots \\geqslant x_{n-1} \\geqslant x_n$.\n记 $v=\\left\\{\\left(x_1, x_2, \\cdots, x_n\\right) \\mid x_i \\geqslant 0,1 \\leqslant i \\leqslant n, \\sum_{i=1}^n x_i=1\\right\\}$.\n令 $F(v)=\\sum_{1 \\leqslant i<j \\leqslant n} x_i x_j\\left(x_i+x_j\\right)=\\sum_{i=1}^n x_i^2-\\sum_{i=1}^n x_i^3$.\n取 $w=\\left\\{\\left(x_1, x_2, \\cdots, x_{n-2}, x_{n-1}+x_n, 0\\right)\\right\\}$, 则\n$$\nF(w)-F(v)=x_{n-1} x_n\\left[2-3\\left(x_{n-1}+x_n\\right)\\right] .\n$$\n又由于 $\\frac{1}{2}\\left(x_{n-1}+x_n\\right) \\leqslant \\frac{1}{n}\\left(x_1+x_2+\\cdots+x_n\\right)$, 故 $x_{n-1}+x_n \\leqslant \\frac{2}{n} \\leqslant \\frac{2}{3}$, 于是 $F(w) \\geqslant F(v)$.\n然后, 将 $x_1, x_2, \\cdots, x_{n-2}, x_{n-1}+x_n$ 从大到小排为 $x_1^{\\prime}, x_2^{\\prime}, \\cdots, x_{n-1}^{\\prime}$, 有 $F(v) \\leqslant F\\left(x_1^{\\prime}, x_2^{\\prime}, \\cdots, x_{n-1}^{\\prime}, 0\\right)$.\n利用上述结论, 又有 $F(v) \\leqslant F\\left(x_1^{\\prime}, x_2^{\\prime}, \\cdots, x_{n-1}^{\\prime}, 0\\right) \\leqslant F\\left(x_1^{\\prime \\prime}, x_2^{\\prime \\prime}, \\cdots\\right.$, $\\left.x_{n-2}^{\\prime \\prime}, 0,0\\right)$, 且 $\\sum_{k=1}^{n-2} x_k^{\\prime \\prime}=1$.\n最后得到 $F(v) \\leqslant f(a, b, 0, \\cdots, 0)=a^2+b^2-a^3-b^3 \\leqslant \\frac{1}{4}$ (这里 $a+ b=1, a, b \\in[0,1]$. ), 等号当 $x_1=x_2=\\frac{1}{2}, x_3=x_4=\\cdots=x_n=0$ 时取到.",6    "remark": "",7    "figures": []8}