高一數論 設p，q，r皆為正實數，求(8q)/(p+3q+r)-(3q+r)/(2p+q+r)-(4p)/(p+2q+r)的最大值為？
回答 (1)
2016-05-01 10:41 am
Sol
p+3q+r=a
2p+q+r=b
p+2q+r=c
=>
p=a+b－2c
q=a－c
r=－3a－b+5c
3q+r=2c－b
(8q)/(p+3q+r)－(3q+r)/(2p+q+r)－(4p)/(p+2q+r)
=(8a－8c)/a－(2c－b)/b－4(a+b－2c)/c
=8a－8c/a－2c/b+1－4a/c－4b/c+8
=17－8c/a－2c/b－4a/c－4b/c
=17－(8c/a+4a/c)－(2c/b+4b/c)
<=17－2*32^(1/2)－2*8^(1/2)
=17－8*2^(1/2)－4*2^(1/2)
=17－12√2
p+3q+r=a
2p+q+r=b
p+2q+r=c
=>
p=a+b－2c
q=a－c
r=－3a－b+5c
3q+r=2c－b
(8q)/(p+3q+r)－(3q+r)/(2p+q+r)－(4p)/(p+2q+r)
=(8a－8c)/a－(2c－b)/b－4(a+b－2c)/c
=8a－8c/a－2c/b+1－4a/c－4b/c+8
=17－8c/a－2c/b－4a/c－4b/c
=17－(8c/a+4a/c)－(2c/b+4b/c)
<=17－2*32^(1/2)－2*8^(1/2)
=17－8*2^(1/2)－4*2^(1/2)
=17－12√2
收錄日期: 2021-04-30 21:30:05
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160430081406AAX3R7O