随手记一道题

2022-06-25 11:06 作者:子瞻Louis 0人读过 | 我要投稿

题目

设 $(a_i)_%7B1%5Cle%20i%5Cle%20n%7D%2C(b_i)_%7B1%5Cle%20i%5Cle%20n%7D$ 是实数， $(c_i)_%7B1%5Cle%20i%5Cle%20n%7D$ 是正实数，求证：

$%5Cleft(%5Csum_%7Bi%3D1%7D%5En%5Csum_%7Bj%3D1%7D%5En%5Cfrac%7Ba_ia_j%7D%7Bc_i%2Bc_j%7D%5Cright)%5Cleft(%5Csum_%7Bi%3D1%7D%5En%5Csum_%7Bj%3D1%7D%5En%5Cfrac%7Bb_ib_j%7D%7Bc_i%2Bc_j%7D%5Cright)%5Cge%5Cleft(%5Csum_%7Bi%3D1%7D%5En%5Csum_%7Bj%3D1%7D%5En%5Cfrac%7Ba_ib_j%7D%7Bc_i%2Bc_j%7D%5Cright)%5E2$

证明：令

$A%3D%5Csum_%7Bi%3D1%7D%5En%5Csum_%7Bj%3D1%7D%5En%5Cfrac%7Ba_ia_j%7D%7Bc_i%2Bc_j%7D%2CB%3D%5Csum_%7Bi%3D1%7D%5En%5Csum_%7Bj%3D1%7D%5En%5Cfrac%7Ba_ib_j%7D%7Bc_i%2Bc_j%7D%2CC%3D%5Csum_%7Bi%3D1%7D%5En%5Csum_%7Bj%3D1%7D%5En%5Cfrac%7Bb_ib_j%7D%7Bc_i%2Bc_j%7D$

再令 $f(x)%3DAx%5E2%2B2Bx%2BC$ ，则其判别式

$%5CDelta(f)%3D4B%5E2-4AC$

因此题目等价于证明 $%5CDelta(f)%5Cle0$ ，也就是 $f(x)%5Cge0$ ，有

$f(x)%3D%5Csum_%7Bi%3D1%7D%5En%5Csum_%7Bj%3D1%7D%5En%5Cfrac%7B(a_ix%2Bb_i)(a_jx%2Bb_j)%7D%7Bc_i%2Bc_j%7D$

令 $x_i%3Da_ix%2Bb_i$ ，于是题目再转化为证明

$%5Csum_%7Bi%3D1%7D%5En%5Csum_%7Bj%3D1%7D%5En%5Cfrac%7Bx_ix_j%7D%7Bc_i%2Bc_j%7D%5Cge0$

置

$g(t)%3D%5Csum_%7Bi%3D1%7D%5En%5Csum_%7Bj%3D1%7D%5En%5Cfrac%7Bx_ix_j%7D%7Bc_i%2Bc_j%7De%5E%7B(c_i%2Bc_j)t%7D$

则有

$g'(t)%3D%5Csum_%7Bi%3D1%7D%5En%5Csum_%7Bj%3D1%7D%5Enx_ix_je%5E%7Bc_it%2Bc_jt%7D%3D%5Cleft(%5Csum_%7Bi%3D1%7D%5En%20x_ie%5E%7Bc_it%7D%5Cright)%5E2%5Cge0$

又有 $g(t)%5Cxrightarrow%7Bt%5Cto-%5Cinfty%7D0$ ，因此

$g(0)%3D%5Csum_%7Bi%3D1%7D%5En%5Csum_%7Bj%3D1%7D%5En%5Cfrac%7Bx_ix_j%7D%7Bc_i%2Bc_j%7D%5Cge0$

Q.E.D.

还有一种别人发的更简单的证法：

证明(2)：由Cauchy不等式

$%5Cint_a%5Eb%7Cf(x)%7C%5E2%5Cmathrm%20d%20x%5Cint_a%5Eb%7Cg(y)%7C%5E2%5Cmathrm%20d%20y%5Cge%5Cleft(%5Cint_a%5Eb%7Cf(x)g(x)%7C%5Cmathrm%20dx%5Cright)%5E2$

可知

$%5Cbegin%7Balign%7D%26%5Cleft(%5Csum_%7Bi%3D1%7D%5En%5Csum_%7Bj%3D1%7D%5En%5Cfrac%7Ba_ia_j%7D%7Bc_i%2Bc_j%7D%5Cright)%5Cleft(%5Csum_%7Bi%3D1%7D%5En%5Csum_%7Bj%3D1%7D%5En%5Cfrac%7Bb_ib_j%7D%7Bc_i%2Bc_j%7D%5Cright)%5C%5C%26%3D%5Cint_0%5E%5Cinfty%5Cleft(%5Csum_%7Bi%3D1%7D%5Ena_ie%5E%7B-c_ix%7D%5Cright)%5E2%5Cmathrm%20dx%5Cint_0%5E%5Cinfty%5Cleft(%5Csum_%7Bi%3D1%7D%5Enb_ie%5E%7B-c_iy%7D%5Cright)%5E2%5Cmathrm%20dy%5C%5C%26%5Cge%5Cleft(%5Cint_0%5E%5Cinfty%5Csum_%7Bi%3D1%7D%5En%5Csum_%7Bj%3D1%7D%5Ena_ib_je%5E%7B-(c_i%2Bc_j)x%7D%5Cmathrm%20dx%5Cright)%5E2%3D%5Cleft(%5Csum_%7Bi%3D1%7D%5En%5Csum_%7Bj%3D1%7D%5En%5Cfrac%7Ba_ib_j%7D%7Bc_i%2Bc_j%7D%5Cright)%5E2%5Cend%7Balign%7D$

Q.E.D

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