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spq法的簡(jiǎn)單運(yùn)用(二)

2023-08-05 12:42 作者:夢(mèng)違Changer  | 我要投稿

我們提升難度,來看例二

例二:非負(fù)實(shí)數(shù)a%E3%80%81b%E3%80%81c滿足ab%2Bbc%2Bca%3D1%0A,求證:%5Cfrac%7B1%2Ba%5E2%20b%5E2%20%20%7D%7B(a%2Bb)%5E2%20%7D%2B%5Cfrac%7B1%2Bb%5E2%20c%5E2%20%20%7D%7B(b%2Bc)%5E2%20%7D%2B%5Cfrac%7B1%2Bc%5E2%20a%5E2%20%20%7D%7B(c%2Ba)%5E2%20%7D%5Cgeq%20%5Cfrac%7B5%7D%7B2%7D%20

證明:左式=%5Cfrac%7B%5Csum_%7Bcyc%7D%5E%7B%7D(1%2Ba%5E2%20b%5E2%20)(a%2Bc)%5E2(b%2Bc)%5E2%7D%7B(a%2Bb)%5E2(b%2Bc)%5E2(c%0A%2Ba)%5E2%20%20%7D%20%20=%5Cfrac%7B%5Csum_%7Bcyc%7D%5E%7B%7D(1%2Ba%5E2%20b%5E2)(1%2Bc%5E2)%5E2%7D%7B(1%2Ba%5E2)(1%2Bb%5E2)(1%2Bc%5E2)%7D%20

故原不等式%5Ciff%202%5Csum_%7Bcyc%7D%5E%7B%7D(1%2Ba%5E2%20b%5E2)(1%2Bc%5E2)%5E2%5Cgeq%205(1%2Ba%5E2)(1%2Bb%5E2)(1%2Bc%5E2)

%5Ciff%206%2B2%5Csum_%7Bcyc%7D%5E%7B%7Da%5E2%20b%5E2%20%2B4%20%5Csum_%7Bcyc%7D%5E%7B%7Da%5E2%20%2B12a%5E2%20b%5E2%20c%5E2%20%2B2%5Csum_%7Bcyc%7D%5E%7B%7Da%5E4%20%2B2a%5E2%20b%5E2%20c%5E2%5Csum_%7Bcyc%7D%5E%7B%7Da%5E2%5Cgeq5%2B5%5Csum_%7Bcyc%7D%5E%7B%7Da%5E2%2B5%5Csum_%7Bcyc%7D%5E%7B%7Da%5E2%20b%5E2%20%2B5a%5E2%20b%5E2%20c%5E2%20%20%20%20%20%5Ciff%201-%5Csum_%7Bcyc%7D%5E%7B%7Da%5E2-3%5Csum_%7Bcyc%7D%5E%7B%7Da%5E2b%5E2%20%2B7a%5E2%20b%5E2%20c%5E2%2B2%5Csum_%7Bcyc%7D%5E%7B%7Da%5E4%2B2a%5E2%20b%5E2%20c%5E2%5Csum_%7Bcyc%7D%5E%7B%7Da%5E2%20%5Cgeq0%20%20%20%20%20

%5Csum_%7Bcyc%7D%5E%7B%7Da%3Ds%E3%80%811%3D%5Csum_%7Bcyc%7D%5E%7B%7Dab%3Dq%E3%80%81abc%3Dp%0A,

%5Ciff%201-(s%5E2%20-2)-3(1-2ps)%2B7p%5E2%2B2(s%5E4-4s%5E2%2B4sp%2B2)%2B2p%5E2%20(s%5E2%20-2)%5Cgeq%200

%5Ciff%202s%5E4%20%2B(2p%5E2-9)s%5E2%2B14ps%2B3p%5E2%2B4%5Cgeq%200%20%20%20%20(*)

我們不妨直接借用例一的證法:使用三次Schur%0A不等式:9p%5Cgeq%0A4s-s%5E3%0A進(jìn)行放縮

那么(*)%E5%B7%A6%E8%BE%B9%5Cgeq%20%5Cfrac%7B2%7D%7B81%7Ds%5E8-%5Cfrac%7B13%7D%7B81%7D%20s%5E6%2B%5Cfrac%7B44%7D%7B81%7D%20s%5E4%20-%5Cfrac%7B59%7D%7B27%7D%20s%5E2%20%2B4

記不等號(hào)右邊的式子為關(guān)于s的函數(shù)g(s),畫出其圖像:

可以看到,對(duì)于s%5Cin%20%5B2%2C%2B%E2%88%9E)%0A,g(s)%5Cgeq%200是我們想要的,但對(duì)于s%5Cin%20%5B%5Csqrt%7B3%7D%20%2C2%5D,我們?nèi)孕栌懻?。不幸的是,這個(gè)討論極為困難,因?yàn)?img type="latex" class="latex" src="https://api.bilibili.com/x/web-frontend/mathjax/tex?formula=(*)" alt="(*)">式不等號(hào)左邊p是作為s的系數(shù)存在,未完全分離,不容易繼續(xù)放縮。

但我們不妨換一種放縮方式,不去放p,而將sp一起放掉,這就需要四次Schur%0A不等式:6sp%5Cgeq%20%20-s%5E4%2B5s%5E2-4?

那么,(*)%5Ciff%20(2s%5E2%2B3)%20p%5E2%2B14sp%5Cgeq%20%209s%5E2-2s%5E4%20-4%20

%E5%B7%A6%E8%BE%B9%5Cgeq%20(2s%5E2%2B3)%20(%5Cfrac%7B1%7D%7B81%7Ds%5E6-%5Cfrac%7B8%7D%7B81%7Ds%5E4%2B%5Cfrac%7B16%7D%7B81%7Ds%5E2)%20%20%20%20%20%20-%5Cfrac%7B7%7D%7B3%7Ds%5E4%2B%5Cfrac%7B35%20%7D%7B3%7D%20s%5E2-%5Cfrac%7B28%7D%7B3%7D

只需證(2s%5E2%2B3)%20(%5Cfrac%7B1%7D%7B81%7Ds%5E6-%5Cfrac%7B8%7D%7B81%7Ds%5E4%2B%5Cfrac%7B16%7D%7B81%7Ds%5E2)%20%20%20%5Cgeq%20%20%20%20%5Cfrac%7B1%7D%7B3%7Ds%5E4-%5Cfrac%7B8%7D%7B3%7D%20s%5E2%2B%5Cfrac%7B16%7D%7B3%7D

%5Ciff%20%5Cfrac%7B2%7D%7B81%7Ds%5E8-%5Cfrac%7B13%7D%7B81%7D%20s%5E6-%5Cfrac%7B19%7D%7B81%7D%20s%5E4%20%2B%5Cfrac%7B88%7D%7B27%7D%20s%5E2%20-%5Cfrac%7B16%20%7D%7B3%7D%20%5Cgeq%200

u(s)為上式不等號(hào)左邊的部分,注意a%E3%80%81b%E3%80%81c%0A非負(fù)且q%3D1,故s%5Cgeq%20%5Csqrt%7B3q%7D%20%3D%5Csqrt%7B3%7D%20

從函數(shù)圖像可以看出u(s)%5B%5Csqrt%7B3%7D%2C%2B%20%E2%88%9E)上恒大于等于0,從而原命題得證,下面我們證明這一斷言:

顯然u(s)是定義在R上的連續(xù)函數(shù),且u(s)%3D0僅有四個(gè)實(shí)根:%5Cpm%202%E3%80%81%5Cpm%20%5Csqrt%7B3%7D%20

u'(s)%3D%5Cfrac%7B16%7D%7B81%7Ds%5E7%20-%5Cfrac%7B78%7D%7B81%7Ds%5E5-%5Cfrac%7B76%7D%7B81%7Ds%5E3%2B%5Cfrac%7B176%7D%7B27%7Ds%20%20%20%20%0A,

u'(s)%3D0有五個(gè)實(shí)根:%5Cpm%202%E3%80%810%E3%80%81%5Cpm%20%5Csqrt%7B%5Cfrac%7B7%2B%5Csqrt%7B2161%20%7D%20%7D%7B16%20%7D%20%20%7D%20

我們只考慮u'(s)%5B%5Csqrt%7B3%7D%2C%2B%20%E2%88%9E)上的取值

u'(%5Csqrt%7B3%7D)%3D%5Cfrac%7B10%20%7D%7B27%7D%5Csqrt%7B3%7D%20%20%20%EF%BC%9E0%EF%BC%8Cu'(%5Csqrt%7B%5Cfrac%7B7%2B%5Csqrt%7B2161%20%7D%20%7D%7B16%20%7D%20%20%7D%20)%3Du'(2)%3Du'(0)%3D0,

注意到u'(s)(0%2C%5Csqrt%7B%5Cfrac%7B7%2B%5Csqrt%7B2161%20%7D%20%7D%7B16%20%7D%20%20%7D%20)上與x軸沒有交點(diǎn),且%5Csqrt%7B3%7D%20%5Cin%20(0%2C%5Csqrt%7B%5Cfrac%7B7%2B%5Csqrt%7B2161%20%7D%20%7D%7B16%20%7D%20%20%7D%20)%E3%80%81u'(%5Csqrt%7B3%7D)%EF%BC%9E0%20

從而u'(s)(0%2C%5Csqrt%7B%5Cfrac%7B7%2B%5Csqrt%7B2161%20%7D%20%7D%7B16%20%7D%20%20%7D%20)上恒大于0

同理可知u'(s)(%20%5Csqrt%7B%5Cfrac%7B7%2B%5Csqrt%7B2161%20%7D%20%7D%7B16%20%7D%20%20%7D%20%2C2%0A)上恒小于0,在(2%2C%2B%20%E2%88%9E)上恒大于0

u(s)(%5Csqrt%7B3%7D%20%2C%5Csqrt%7B%5Cfrac%7B7%2B%5Csqrt%7B2161%20%7D%20%7D%7B16%20%7D%20%20%7D%20)上單調(diào)遞增,在(%20%5Csqrt%7B%5Cfrac%7B7%2B%5Csqrt%7B2161%20%7D%20%7D%7B16%20%7D%20%20%7D%20%2C2%0A)上單調(diào)遞減,在(2%2C%2B%20%E2%88%9E)上單調(diào)遞增

注意u(%5Csqrt%7B3%7D%20)%3Du(2)%3D0,從而u(s)%5Cgeq%200,當(dāng)且僅當(dāng)s%3D%5Csqrt%7B3%7D%20%E6%88%962時(shí)取等

s%3D%5Csqrt%7B3%7D意味著a%3Db%3Dc,與Schur不等式的取等條件矛盾

故僅有s%3D2時(shí)等號(hào)成立,此時(shí)(a%2Cb%2Cc)%3D(1%2C1%2C0)(1%2C0%2C1)(0%2C1%2C1)

至此,原命題得證!


spq法的簡(jiǎn)單運(yùn)用(二)的評(píng)論 (共 條)

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