✔ 最佳答案
a,b和c為正實數,且abc=1,證明:
1/[(a^2)*(b+c)]+1/[(b^2)*(c+a)]+1/[(c^2)*(a+b)] >=3/2
Solution:
利用 AM >= GM :
1/[(a^2)*(b+c)]+1/[(b^2)*(c+a)]+1/[(c^2)*(a+b)]
>=
3 * { 1/[(a^2)*(b+c)] * 1/[(b^2)*(c+a)] * 1/[(c^2)*(a+b)] } ^ (1/3)
= 3 * { 1 / [ (abc)^2 * (b+c)(c+a)(a+b) ] ^ (1/3)
= 3 * [ 1 / (b+c)(c+a)(a+b)] ^ (1/3)
>= 3 * [ 1 / (2√bc)(2√ca)(2√ab)] ^ (1/3)
= 3 * [ 1 / 8√(abc)^2] ^ (1/3)
= 3 * (1/8)^(1/3)
= 3/2
證畢
2009-10-23 11:20:12 補充:
Sorry! 上面 3 * [ 1 / (b+c)(c+a)(a+b)] ^ (1/3) >= 3 * [ 1 / (2√bc)(2√ca)(2√ab)] ^ (1/3)
是錯的。下面的證明才是對的 :
a,b和c為正實數,且abc=1,證明:
1/[(a^2)*(b+c)]+1/[(b^2)*(c+a)]+1/[(c^2)*(a+b)] >=3/2
**********************************************************************************************
2009-10-23 11:21:30 補充:
由abc = 1 , 即證 :
abc/[(a^2)*(b+c)] + abc/[(b^2)*(c+a)] + abc/[(c^2)*(a+b)] >=3/2
即證 :bc/(ab + ac) + ac/(bc + ab) + ab/(ac + bc) >= 3/2
即證 :1 + [bc/(ab + ac)] + 1 + [c/(bc + ab)] + 1 + [ab/(ac + bc)] >= 3/2 + 3
即證 :(ab + bc + ac) [1/(ab + ac) + 1/(bc + ab) + 1/(ac + bc)] >= 9/2
2009-10-23 11:23:45 補充:
即證 :2(ab + bc + ac) [1/(ab + ac) + 1/(bc + ab) + 1/(ac + bc)] >= 9
即證 :[(ab+ac) + (bc+ab) + (ac+bc)] [1/(ab + ac) + 1/(bc + ab) + 1/(ac + bc)] >= 9
用AM >= GM :
2009-10-23 11:24:09 補充:
[(ab+ac) + (bc+ab) + (ac+bc)] [1/(ab + ac) + 1/(bc + ab) + 1/(ac + bc)]
>= {3[(ab+ac) (bc+ab) (ac+bc)] ^(1/3)} * 3 / {[(ab+ac) (bc+ab) (ac+bc)] ^(1/3)}
= 9 , 成立!
因此 1/[(a^2)*(b+c)]+1/[(b^2)*(c+a)]+1/[(c^2)*(a+b)] >=3/2 成立!
證畢。