\u4ee5\u4e0b\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n
(1)<\/strong>\u3000\\(n\\) \u3092\u81ea\u7136\u6570, \\(a\\) \u3092\u6b63\u306e\u5b9a\u6570\u3068\u3057\u3066,\r\n\\[\r\nf(x) = (n+1) \\left\\{ \\log (a+x) -\\log (n+1) \\right\\} -n \\left( \\log a -\\log n \\right) -\\log x\r\n\\]\r\n\u3068\u304a\u304f. \\(x \\gt 0\\) \u306b\u304a\u3051\u308b\u95a2\u6570 \\(f(x)\\) \u306e\u6975\u5024\u3092\u6c42\u3081\u3088. \u305f\u3060\u3057, \u5bfe\u6570\u306f\u81ea\u7136\u5bfe\u6570\u3068\u3059\u308b.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(n\\) \u304c \\(2\\) \u4ee5\u4e0a\u306e\u81ea\u7136\u6570\u306e\u3068\u304d, \u6b21\u306e\u4e0d\u7b49\u5f0f\u304c\u6210\u308a\u7acb\u3064\u3053\u3068\u3092\u793a\u305b.\r\n\\[\r\n\\dfrac{1}{n} \\textstyle\\sum\\limits _ {k=1}^n \\dfrac{k+1}{k} \\gt (n+1)^{\\frac{1}{n}}\r\n\\]<\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \\[\\begin{align}\r\nf'(x) & = \\dfrac{n+1}{a+x} -\\dfrac{1}{x} \\\\\r\n& = \\dfrac{(n+1) x -(a+x)}{x (a+x)} \\\\\r\n& = \\dfrac{nx -a}{x (a+x)} \\ .\r\n\\end{align}\\]\r\n\\(f'(x) = 0\\) \u3092\u3068\u304f\u3068\r\n\\[\r\nx =\\dfrac{a}{n} \\ .\r\n\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(f(x)\\) \u306e\u5897\u6e1b\u306f\u4e0b\u8868\u306e\u3088\u3046\u306b\u306a\u308b.\r\n\\[\r\n\\begin{array}{c|cccc} x & (0) & \\cdots & \\dfrac{a}{n} & \\cdots \\\\ \\hline f'(x) & & - & 0 & + \\\\ \\hline f(x) & & \\searrow & \\text{\u6975\u5c0f} & \\nearrow \\end{array}\r\n\\]\r\n\u3088\u3063\u3066, \u6c42\u3081\u308b\u6975\u5024\u306f\r\n\\[\\begin{align}\r\nf \\left( \\dfrac{a}{n} \\right) & = (n+1) \\left\\{ \\log \\dfrac{a (n+1)}{n} -\\log (n+1) \\right\\} \\\\\r\n& \\qquad -n \\log \\dfrac{a}{n} -\\log \\dfrac{a}{n} \\\\\r\n& = \\underline{0} \\ .\r\n\\end{align}\\]\r\n (2)<\/strong><\/p>\r\n (1)<\/strong> \u306e\u7d50\u679c\u3088\u308a, \\(f(x) \\geqq 0\\) \u306a\u306e\u3067\r\n\\[\\begin{align}\r\n\\log \\left( \\dfrac{a+x}{n+1} \\right)^{n+1} -\\log & \\left( \\dfrac{a}{n} \\right)^n x \\geqq 0 \\\\\r\n\\text{\u2234} \\quad \\left( \\dfrac{a+x}{n+1} \\right)^{n+1} & \\geqq \\left( \\dfrac{a}{n} \\right)^n x \\quad ... [1] \\ .\r\n\\end{align}\\]\r\n\u793a\u3057\u305f\u3044\u4e0d\u7b49\u5f0f\r\n\\[\r\n\\dfrac{1}{n} \\textstyle\\sum\\limits _ {k=1}^n \\dfrac{k+1}{k} \\gt (n+1)^{\\frac{1}{n}} \\quad ... [ \\text{A} ]\r\n\\]\r\n\u304c, \\(2\\) \u4ee5\u4e0a\u306e\u81ea\u7136\u6570 \\(n\\) \u306b\u3064\u3044\u3066\u6210\u7acb\u3059\u308b\u3053\u3068\u3092, \u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u3092\u7528\u3044\u3066\u793a\u3059.<\/p>\r\n 1*<\/strong>\u3000\\(n = 2\\) \u306e\u3068\u304d\r\n\\[\\begin{align}\r\n\\dfrac{1}{2} \\left( \\dfrac{2}{1} +\\dfrac{3}{2} \\right) & = \\dfrac{7}{4} \\ , \\\\\r\n( 2 +1 )^{\\frac{1}{2}} & = \\sqrt{3} \\ .\r\n\\end{align}\\]\r\n\\(\\left( \\dfrac{7}{4} \\right)^2 = \\dfrac{49}{16} \\gt 3\\) \u306a\u306e\u3067, [A] \u304c\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n 2*<\/strong>\u3000\\(n = m \\ ( m \\geqq 2 )\\) \u306e\u3068\u304d \u4ee5\u4e0a\u3088\u308a, \u984c\u610f\u306f\u793a\u3055\u308c\u305f.<\/p>\r\n","protected":false},"excerpt":{"rendered":"\u4ee5\u4e0b\u306e\u554f\u3044\u306b\u7b54\u3048\u3088. (1)\u3000\\(n\\) \u3092\u81ea\u7136\u6570, \\(a\\) \u3092\u6b63\u306e\u5b9a\u6570\u3068\u3057\u3066, \\[ f(x) = (n+1) \\left\\{ \\log (a+x) -\\log (n+1) \\right\\} -n \\left( \\ … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n
\r\n
\r\n[A] \u304c\u6210\u7acb\u3059\u308b\u3068\u4eee\u5b9a\u3059\u308b\u3068\r\n\\[\\begin{align}\r\n\\dfrac{1}{m} \\textstyle\\sum\\limits _ {k=1}^m \\dfrac{k+1}{k} & \\gt (m+1)^{\\frac{1}{m}} \\\\\r\n\\text{\u2234} \\quad \\textstyle\\sum\\limits _ {k=1}^m \\dfrac{k+1}{k} & \\gt m (m+1)^{\\frac{1}{m}} \\ .\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \u3053\u306e\u3068\u304d\r\n\\[\r\n\\left( \\dfrac{1}{m+1} \\textstyle\\sum\\limits _ {k=1}^{m+1} \\dfrac{k+1}{k} \\right)^{m+1} \\gt \\left\\{ \\dfrac{m (m+1)^{\\frac{1}{m}} +\\frac{m+2}{m+1}}{m+1} \\right\\}^{m+1} \\quad ... [2] \\ .\r\n\\]\r\n\u3053\u3053\u3067, \\(a = m (m+1)^{\\frac{1}{m}}\\) , \\(x = \\dfrac{m+2}{m+1}\\) , \\(n = m\\) \u3068\u304a\u3044\u3066, [1] \u3092\u7528\u3044\u308c\u3070\r\n\\[\r\n[2] \\geqq \\left\\{ \\dfrac{m (m+1)^{\\frac{1}{m}}}{m} \\right\\}^m \\dfrac{m+2}{m+1} = m+2 \\ .\r\n\\]\r\n\u3057\u305f\u304c\u3063\u3066\r\n\\[\r\n\\dfrac{1}{m+1} \\textstyle\\sum\\limits _ {k=1}^{m+1} \\dfrac{k+1}{k} \\gt (m+2)^{\\frac{1}{m+1}} \\ .\r\n\\]\r\n\u3086\u3048\u306b, \\(n = m+1\\) \u306e\u3068\u304d\u3082 [A] \u304c\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n<\/ol>\r\n