[PURE MATHS]inequalities2條[10分]－貨運百科

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) <= ((a1)^2+(a2)^2+ (a3)^2)^(1/2) +... 顯示更多 1.a,b,c>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) <= ((a1)^2+(a2)^2+ (a3)^2)^(1/2) + ((b1)^2+(b2)^2+(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 <=(a1^2+a2^2+a3^2)(b1^2+b2^2+b3^2)(a1b1+a2b2+a3b3) <=[(a1^2+a2^2+a3^2)(b1^2+b2^2+b3^2)]^(1/2) 2008-01-01 23:28:19 補充： (a1+b1)^2 + (a2+b2)^2 +(a3+b3)^2 =(a1)^2+(a2)^2+ (a3)^2+ (b1)^2+(b2)^2+(b3)^2+2(a1b1+a2b2+a3b3)<=((a1)^2+(a2)^2+ (a3)^2) + (b1)^2+(b2)^2+(b3)^2+2[(a1^2+a2^2+a3^2)(b1^2+b2^2+b3^2)]^(1/2)=[(a1^2+a2^2+a3^2)^(1/2)+(b1^2+b2^2+b3^2)^(1/2)]^2Taking square root ,u will get the answer