09/09/20 00:24:25
>>462
(略証)
問題の左辺をＳとおく。 [初代スレ.501] より
a/(b+c) + 2b/(c+d) - {(a+b)/(b+c) + (b+c)/(c+d) -1} = (b^2 +cd)/[(b+c)(c+d)],
b/(c+d) + 2c/(d+e) - {(b+c)/(c+d) + (c+d)/(d+e) -1} ＞ (c^2)/[(c+d)(d+e)],
それぞれ 0.3 と 0.7 を掛けて加えると、
0.3a/(b+c) + 1.3b/(c+d) + 1.4c/(d+e) - {0.3(a+b)/(b+c) + (b+c)/(c+d) + 0.7(c+d)/(d+e) -1}
> 0.3(b^2 +cd)/[(b+c)(c+d)] + 0.7(c^2)/[(c+d)(d+e)]
> {0.3(b^2 +cd)d + 0.7(b+c)c^2}/[(b+c)(c+d)(d+e)]
> {(0.2b^2 + 0.3cd)d + (0.4b + 0.7c)c^2}/[(b+c)(c+d)(d+e)]
= {0.4c(b+c)(c+d) + 0.2(b-c)^2･d + 0.3c(c-d)^2}/[(b+c)(c+d)(d+e)]
> 0.4c(b+c)(c+d)/[(b+c)(c+d)(d+e)]
= 0.4c/(d+e),
∴ 0.3a/(b+c) + 1.3b/(c+d) + c/(d+e) ＞ 0.3(a+b)/(b+c) + (b+c)/(c+d) + 0.7(c+d)/(d+e) -1,
循環的に加えて
2.6Ｓ > (0.3 + 1 + 0.7)Σ[k=1,n] (a_k +a_{k+1})/(a_{k+1} +a_{k+2}) - n
> (0.3 + 1 + 0.7)n - n (← 相加・相乗平均)
= 2n - n = n.
∴ Ｓ > n/2.6
ぬるぽ
･Shapiro 巡回不等式 関連レス
[第２章.284-285]
[第３章.172-173, 218-220]