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一些有趣的數(shù)學(xué)題 (Day 1)

2022-05-18 11:10 作者:中華最菜蒟蒻OIer  | 我要投稿

已知 %5Cdfrac%7B1%7D%7Be%5E2%7D%3E%5Cln%5Cleft(%5Cln%5Cpi%5Cright),求證:%5Cdfrac%7B%5Cln%5Cleft(e%5E2%2B1%5Cright)-2%7D%7B%5Cln%5Cleft(%5Cln(%5Cln%5Cpi)%2B1%5Cright)%7D%2B%5Cdfrac%7B%5Cln%5Cleft(%5Cln(%5Cln%5Cpi)%2B1%5Cright)%7D%7B%5Cln%5Cleft(e%5E2%2B1%5Cright)-2%7D%3E%5Cdfrac%7B%5Cleft(2e%5E2%2B1%5Cright)%5Cln%5Cleft(%5Cln%5Cpi%5Cright)%7D%7B2%2B%5Cln%5Cleft(%5Cln%5Cpi%5Cright)%7D%2B%5Cdfrac%7B2%2B%5Cln%5Cleft(%5Cln%5Cpi%5Cright)%7D%7B%5Cleft(2e%5E2%2B1%5Cright)%5Cln%5Cleft(%5Cln%5Cpi%5Cright)%7D。

解法會(huì)在下面給出,大家可以選擇先自己思考一下這個(gè)問題。

證明:注意到該不等式兩邊都是形如 %5Cdfrac%7Ba%7D%7Bb%7D%2B%5Cdfrac%7Bb%7D%7Ba%7D?的形式,故可設(shè)t_1%3D%5Cdfrac%7B%5Cln%5Cleft(e%5E2%2B1%5Cright)-2%7D%7B%5Cln%5Cleft(%5Cln(%5Cln%5Cpi)%2B1%5Cright)%7D%2Ct_2%3D%5Cdfrac%7B2%2B%5Cln%5Cleft(%5Cln%5Cpi%5Cright)%7D%7B%5Cleft(2e%5E2%2B1%5Cright)%5Cln%5Cleft(%5Cln%5Cpi%5Cright)%7D,原不等式即化為 t_1%2B%5Cdfrac%7B1%7D%7Bt_1%7D%3Et_2%2B%5Cdfrac%7B1%7D%7Bt_2%7D。

考慮 x%2B%5Cdfrac%7B1%7D%7Bx%7D(x%3E0)?的單調(diào)性,易知它在 (0%2C1%5D?上單調(diào)遞減,在 (1%2C%2B%5Cinfty)?上單調(diào)遞增,并在 x%3D1?時(shí)取得最小值。接下來我們研究 t_1和?t_2的大小關(guān)系。

對于 t_1,有:t_1%3D%5Cdfrac%7B%5Cln%5Cleft(e%5E2%2B1%5Cright)-2%7D%7B%5Cln%5Cleft(%5Cln(%5Cln%5Cpi)%2B1%5Cright)%7D%3D%5Cdfrac%7B%5Cln%5Cleft(%5Cdfrac%7B1%7D%7Be%5E2%7D%2B1%5Cright)%7D%7B%5Cln%5Cleft(%5Cln(%5Cln%5Cpi)%2B1%5Cright)%7D%3E1;

而對于 t_2,設(shè)函數(shù) f(x)%3D%5Cdfrac%7B2%2Bx%7D%7B%5Cleft(2e%5E2%2B1%5Cright)x%7D,則 t_2%3Df(%5Cln%5Cleft(%5Cln%5Cpi%5Cright)),而且 f%5Cleft(%5Cdfrac%7B1%7D%7Be%5E2%7D%5Cright)%3D1。因?yàn)?f(x)%3D%5Cdfrac%7B2%2Bx%7D%7B%5Cleft(2e%5E2%2B1%5Cright)x%7D%3D%5Cdfrac%7B%5Cdfrac%7B2%7D%7Bx%7D%2B1%7D%7B2e%5E2%2B1%7D,易知 f(x)?在 (0%2C%2B%5Cinfty)?上單調(diào)遞減,所以有 t_2%3Df(%5Cln%5Cleft(%5Cln%5Cpi%5Cright))%3Ef%5Cleft(%5Cdfrac%7B1%7D%7Be%5E2%7D%5Cright)%3D1。

因?yàn)?t_1%2Ct_2%3E1,所以原不等式化為 t_1%3Et_2,即:

%5Cbegin%7Baligned%7D%5Cdfrac%7B%5Cln%5Cleft(e%5E2%2B1%5Cright)-2%7D%7B%5Cln%5Cleft(%5Cln(%5Cln%5Cpi)%2B1%5Cright)%7D%26%3E%5Cdfrac%7B2%2B%5Cln%5Cleft(%5Cln%5Cpi%5Cright)%7D%7B%5Cleft(2e%5E2%2B1%5Cright)%5Cln%5Cleft(%5Cln%5Cpi%5Cright)%7D%5C%5C(2e%5E2%2B1)%5Cln%5Cleft(%5Cdfrac%7B1%7D%7Be%5E2%7D%2B1%5Cright)%26%3E%5Cleft(%5Cdfrac%7B2%7D%7B%5Cln%5Cleft(%5Cln%5Cpi%5Cright)%7D%2B1%5Cright)%5Cln%5Cleft(%5Cln(%5Cln%5Cpi)%2B1%5Cright)%5C%5C%5Cend%7Baligned%7D

設(shè)函數(shù) g%5Cleft(%5Cdfrac%7B1%7D%7Be%5E2%7D%5Cright)%3Eg(%5Cln%5Cleft(%5Cln%5Cpi%5Cright)),則上面的不等式化為 g%5Cleft(%5Cdfrac%7B1%7D%7Be%5E2%7D%5Cright)%3Eg(%5Cln%5Cleft(%5Cln%5Cpi%5Cright))。接下來我們分析 g(x)?的單調(diào)性。

g(x) 求導(dǎo),可得 g'(x)%3D-%5Cdfrac%7B2%5Cln(x%2B1)%7D%7Bx%5E2%7D%2B%5Cdfrac%7Bx%2B2%7D%7Bx(x%2B1)%7D%3D%5Cdfrac%7Bx%5E2%2B2x-2(x%2B1)%5Cln(x%2B1)%7D%7Bx%5E2(x%2B1)%7D。當(dāng) x%3E0?時(shí)有 x%5E2(x%2B1)%3E0,所以我們設(shè) h(x)%3Dx%5E2%2B2x-2(x%2B1)%5Cln(x%2B1),則 h'(x)%3D2x%2B2-(2%5Cln(x%2B1)%2B2)%3D2x-2%5Cln(x%2B1)。因?yàn)?h(0)%3D0h(0)%3D0,而易證 x%3E0?時(shí)有 h'(x)%3E0?恒成立,所以有 h(x)%3E0%2Cg'(x)%3E0?在 x%3E0?時(shí)恒成立,也就是 g(x)?在 (0%2C%2B%5Cinfty)?上單調(diào)遞增。? ?

因?yàn)?%5Cdfrac%7B1%7D%7Be%5E2%7D%3E%5Cln%5Cleft(%5Cln%5Cpi%5Cright)%3E0,所以 g%5Cleft(%5Cdfrac%7B1%7D%7Be%5E2%7D%5Cright)%3Eg(%5Cln%5Cleft(%5Cln%5Cpi%5Cright))。原命題得證。

補(bǔ)充:對 h'(x)%3D2x-2%5Cln(x%2B1)%3E0?在 x%3E0?時(shí)恒成立的證明。? ?

首先證明 e%5Ex%3Ex%2B1?在 x%3E0?時(shí)恒成立。設(shè) p(x)%3De%5Ex-x-1,則 p'(x)%3De%5Ex-1。令 p'(x)%3D0?可得 x%3D0。顯然 p'(x)?在 %5Cmathbb%7BR%7D?上單調(diào)遞增,所以 p'(x)%3E0?在 x%3E0?時(shí)恒成立。又因?yàn)?p(0)%3D0,所以 p(x)%3E0?在 x%3E0?時(shí)恒成立,也即 e%5Ex%3Ex%2B1?在 x%3E0?時(shí)恒成立。? ?

有了這個(gè)式子,則 e%5Ex%3Ex%2B1%5CLongrightarrow%20x%3E%5Cln(x%2B1)%5CLongrightarrow2x-2%5Cln(x%2B1)%3E0%5CLongrightarrow%20h'(x)%3E0。




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