標題:
[PURE MATHS]inequalities2條[10分]
發問:
1.a,b,c>0, ab+bc+ca=1prove:[1+abc(a+b+c)](a+b+c)^2≧42.Let a1 ,a2 ,a3 and b1 b2 b3 be two sets of non negative real numbers.Using the Schwarz's inequality , show that ( (a1+b1)^2 + (a2+b2)^2 +(a3+b3)^2 )^(1/2) 0, ab+bc+ca=1 prove: [1+abc(a+b+c)](a+b+c)^2≧4 2.Let a1 ,a2 ,a3 and b1 b2 b3 be two sets of non negative real numbers. Using the Schwarz's inequality , show that ( (a1+b1)^2 + (a2+b2)^2 +(a3+b3)^2 )^(1/2)
最佳解答:
(A+B+C)^2 =A^2+B^2+C^2+2(AB+BC+CA) =(a^2+b^2+c^2)+2 >=2 Since a,b,c>0, so a^2+b^2+c^2>0 Also, 1+abc(a+b+c) >1+1 =2 Multiply them together , [1+abc(a+b+c)](a+b+c)^2≧4 b)I will think later plz 2008-01-01 23:28:00 補充: 2) I can do it now , By Swartz Inequlity ,(a1b1+a2b2+a3b3)^2
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