兩道三元不等式

2023-05-17 06:50 作者:桌游小黃鴨 0人讀過 | 我要投稿

條件：非負(fù)實(shí)數(shù) $a%2Cb%2Cc$ 滿足 $ab%2Bbc%2Bca%3D1$ ，求證

$%5Cfrac%7B1%7D%7Ba%2Bb%7D%2B%5Cfrac%7B1%7D%7Bb%2Bc%7D%2B%5Cfrac%7B1%7D%7Bc%2Ba%7D%5Cgeq%5Cfrac%7B5%7D%7B2%7D%2C%5C%5C%0A%5Cfrac%7B1%7D%7B(a%2Bb)%5E2%7D%2B%5Cfrac%7B1%7D%7B(b%2Bc)%5E2%7D%2B%5Cfrac%7B1%7D%7B(c%2Ba)%5E2%7D%5Cgeq%5Cfrac%7B9%7D%7B4%7D.$

解答：根據(jù)對(duì)稱性，不妨設(shè) $a%5Cleq%20b%5Cleq%20c$ ，不難判斷出 $a%5Cleq%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D$ . 此時(shí)

$1%2Ba%5E2%3D(a%2Bb)(a%2Bc)%5Cleq%5Cfrac%7B(a%2Bb%2Ba%2Bc)%5E2%7D%7B4%7D$ 可推出 $b%2Bc%5Cgeq%202(%5Csqrt%7Ba%5E2%2B1%7D-a%20)%5Cequiv%202t%2C$

先考察第一小問.?

$%5Cfrac%7B1%7D%7Ba%2Bb%7D%2B%5Cfrac%7B1%7D%7Bb%2Bc%7D%2B%5Cfrac%7B1%7D%7Bc%2Ba%7D%3D%5C%5C%0A%3D%5Cfrac%7Ba%2Bc%2Ba%2Bb%7D%7B(a%2Bb)(a%2Bc)%7D%2B%5Cfrac%7B1%7D%7Bb%2Bc%7D%3D%5Cfrac%7B2a%2B(b%2Bc)%7D%7Ba%5E2%2B1%7D%2B%5Cfrac%7B1%7D%7Bb%2Bc%7D%5Cequiv%5Cfrac%7B2a%2Bx%7D%7Ba%5E2%2B1%7D%2B%5Cfrac%7B1%7D%7Bx%7D%5Cequiv%20f(x).$

計(jì)算導(dǎo)數(shù)可知 $f'(x)%5Cgeq%200%2C%20%5Cforall%20x%5Cgeq%20%5Csqrt%7Ba%5E2%2B1%7D$ ，再利用 $b%2Bc%5Cgeq%202(%5Csqrt%7Ba%5E2%2B1%7D-a%20)%5Cgeq%20%5Csqrt%7Ba%5E2%2B1%7D$ （最后一個(gè)不等式不難由 $a%5Cleq%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D$ 判斷出），

原式左側(cè)= $f(b%2Bc)%5Cgeq%20f(2t)%2C$ 為證明第一問，只需證明在 $t%5E2%2B2at%3D1%2C%20a%5Cleq%20%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D$ 的條件下有

$%5Cfrac%7B2%7D%7Ba%2Bt%7D%2B%5Cfrac%7B1%7D%7B2t%7D%5Cgeq%20%5Cfrac%7B5%7D%7B2%7D$ 即可. 代入 $a%3D%5Cfrac%7B1%7D%7B2t%7D-%5Cfrac%7Bt%7D%7B2%7D$ 可化為 $%5Cfrac%7B9t%5E2%2B1%7D%7B2t(1%2Bt%5E2)%7D%5Cgeq%5Cfrac%7B5%7D%7B2%7D%2C$ 這等價(jià)于 $1-5t%2B9t%5E2-5t%5E3%5Cgeq%200$ ，也即等價(jià)于 $(1-t)((1-2t)%5E2%2Bt%5E2)%5Cgeq%200%2C$ 而這是顯然的（注意 $2at%2Bt%5E2%3D1%2C%20t%5E2%5Cleq1%2Ct%5Cleq1$ ）. 這就說明了原式左側(cè) $%3Df(b%2Bc)%5Cgeq%20f(2t)%5Cgeq%5Cfrac%7B5%7D%7B2%7D.$

然后考慮第二小問.

? $%5Cfrac%7B1%7D%7B(a%2Bb)%5E2%7D%2B%5Cfrac%7B1%7D%7B(b%2Bc)%5E2%7D%2B%5Cfrac%7B1%7D%7B(c%2Ba)%5E2%7D%3D%5C%5C%0A%3D%5Cfrac%7B2a%5E2%2B2a(b%2Bc)%2Bb%5E2%2Bc%5E2%7D%7B((a%2Bb)(a%2Bc))%5E2%7D%2B%5Cfrac%7B1%7D%7B(b%2Bc)%5E2%7D%3D%5C%5C%0A%3D%5Cfrac%7B2a%5E2-2%2B4a(b%2Bc)%2B(b%2Bc)%5E2%7D%7B(a%5E2%2B1)%5E2%7D%2B%5Cfrac%7B1%7D%7B(b%2Bc)%5E2%7D%5Cequiv%5C%5C%0A%5C%5C%20%5Cequiv%20%5Cfrac%7B2a%5E2-2%2B4ax%2Bx%5E2%7D%7B(a%5E2%2B1)%5E2%7D%2B%5Cfrac%7B1%7D%7Bx%5E2%7D%5Cequiv%20g(x).$

求導(dǎo)可知

$g'(x)%3D%5Cfrac%7B4a%2B2x%7D%7B(a%5E2%2B1)%5E2%7D-2x%5E%7B-3%7D%5Cgeq%20%5Cfrac%7B4a%2B4(%5Csqrt%7Ba%5E2%2B1%7D-a)%7D%7B(a%5E2%2B1)%5E2%7D-2x%5E%7B-3%7D%3D%5C%5C%0A%3D%5Cfrac%7B4%7D%7B(%5Csqrt%7Ba%5E2%2B1%7D)%5E3%7D-%5Cfrac%7B2%7D%7Bx%5E3%7D%5Cgeq%20%5Cfrac%7B4%7D%7B(%5Csqrt%7Ba%5E2%2B1%7D)%5E3%7D-%5Cfrac%7B2%7D%7B(%5Csqrt%7Ba%5E2%2B1%7D)%5E3%7D%5Cgeq%200%2C$

對(duì)于 $x%5Cgeq%202(%5Csqrt%7Ba%5E2%2B1%7D-a)$ 成立，其中上式利用了 $2(%5Csqrt%7Ba%5E2%2B1%7D-a%20)%5Cgeq%20%5Csqrt%7Ba%5E2%2B1%7D$ .

仍然令 $t%5Cequiv(%5Csqrt%7Ba%5E2%2B1%7D-a%20)%2C$ 我們有原式左側(cè)= $g(b%2Bc)%5Cgeq%20g(2t)%2C$ 從而只需要在 $t%5E2%2B2at%3D1%2C%20a%5Cleq%20%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D$ 的條件下證明 $%5Cfrac%7B2%7D%7B(a%2Bt)%5E2%7D%2B%5Cfrac%7B1%7D%7B4t%5E2%7D%5Cgeq%20%5Cfrac%7B9%7D%7B4%7D$ 即可。同樣代入 $a%3D%5Cfrac%7B1%7D%7B2t%7D-%5Cfrac%7Bt%7D%7B2%7D$ 并通分，將其等價(jià)化為 $%5Cfrac%7B33t%5E4%2B2t%5E2%2B1%7D%7B(t%5E2%2B1)%5E2t%5E2%7D%5Cgeq%209%2C$ 也即等價(jià)于 $1-7t%5E2%2B15t%5E4-9t%5E6%5Cgeq%200%2C$ 也即等價(jià)于 $(1-t%5E2)(1-3t%5E2)%5E2%5Cgeq%200%2C$ 而這對(duì)于滿足條件的