若a,b,c>0,求证:a三次/(b+c)+b三次/(a+c)+c三次/(a+b)≥1/2(ab+bc+ac)
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若a,b,c>0,求证:a三次/(b+c)+b三次/(a+c)+c三次/(a+b)≥1/2(ab+bc+ac)
若a,b,c>0,求证:a三次/(b+c)+b三次/(a+c)+c三次/(a+b)≥1/2(ab+bc+ac)
若a,b,c>0,求证:a三次/(b+c)+b三次/(a+c)+c三次/(a+b)≥1/2(ab+bc+ac)
证明:利用柯西不等式a三次/(b+c)+b三次/(a+c)+c三次/(a+b)
>=(a^2+b^2+c^2)^2/(a(b+c)+b(a+c)+c(a+b))
= (a^2+b^2+c^2)^2/(2(ab+bc+ac)) ------(1)
显然由(a-b)^2+(b+c)^2+(a+c)^2>=0 =>a^2+b^2+c^2>ab+bc+ac
故(1)右边>=(ab+bc+ac)^2/(2(ab+bc+ac))=1/2(ab+bc+ac)
当且仅当a=b=c时等号成立
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证明:由柯西不等式得a三次/(b+c)+b三次/(a+c)+c三次/(a+b)
>=(a^2+b^2+c^2)^2/(a(b+c)+b(a+c)+c(a+b))
= (a^2+b^2+c^2)^2/(2(ab+bc+ac)) ------(1)
故:(a-b)^2+(b+c)^2+(a+c)^2>=0 =>a^2+b^2+c^2>a...
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证明:由柯西不等式得a三次/(b+c)+b三次/(a+c)+c三次/(a+b)
>=(a^2+b^2+c^2)^2/(a(b+c)+b(a+c)+c(a+b))
= (a^2+b^2+c^2)^2/(2(ab+bc+ac)) ------(1)
故:(a-b)^2+(b+c)^2+(a+c)^2>=0 =>a^2+b^2+c^2>ab+bc+ac
所以(1)右边>=(ab+bc+ac)^2/(2(ab+bc+ac))=1/2(ab+bc+ac)
当且仅当a=b=c时等号成立
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