已知a,b,c是正数,求证a^2a*b^2b*c^2c>=a^(b+c)*b^(c+a)*c^(a+b)
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![已知a,b,c是正数,求证a^2a*b^2b*c^2c>=a^(b+c)*b^(c+a)*c^(a+b)](/uploads/image/z/2528417-65-7.jpg?t=%E5%B7%B2%E7%9F%A5a%2Cb%2Cc%E6%98%AF%E6%AD%A3%E6%95%B0%2C%E6%B1%82%E8%AF%81a%5E2a%2Ab%5E2b%2Ac%5E2c%3E%3Da%5E%28b%2Bc%29%2Ab%5E%28c%2Ba%29%2Ac%5E%28a%2Bb%29)
已知a,b,c是正数,求证a^2a*b^2b*c^2c>=a^(b+c)*b^(c+a)*c^(a+b)
已知a,b,c是正数,求证a^2a*b^2b*c^2c>=a^(b+c)*b^(c+a)*c^(a+b)
已知a,b,c是正数,求证a^2a*b^2b*c^2c>=a^(b+c)*b^(c+a)*c^(a+b)
这道题是《不等式选讲》里的习题吧,答案见这里:
http://hi.baidu.com/%CC%EC%CF%C2%BB%E1%CE%DE%C3%FB/album/item/60a043444902fd0fcefca35f.html#IMG=60a043444902fd0fcefca35f
证明 不妨设a≥b≥c>0,则
(a^(2a)*b^(2b)*c^(2c))/(a^(b+c)*b^(c+a)*c^(a+b))
=(a^a*b^b*c^c)/(a^((b+c)/2)*b^((c+a)/2)*c^((a+b)/2))
=(a^((a-b)/2+(a-c)/2))*(b^((b-c)/2+(b-a)/2))*(c^((c-a)/2+(c-b)/2))
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证明 不妨设a≥b≥c>0,则
(a^(2a)*b^(2b)*c^(2c))/(a^(b+c)*b^(c+a)*c^(a+b))
=(a^a*b^b*c^c)/(a^((b+c)/2)*b^((c+a)/2)*c^((a+b)/2))
=(a^((a-b)/2+(a-c)/2))*(b^((b-c)/2+(b-a)/2))*(c^((c-a)/2+(c-b)/2))
=((a/b)^((a-b)/2))*((a/c)^((a-c)/2))*((b/c)^((b-c)/2))≥1
故得
a^(2a)b^(2b)c^2(2c)≥a^(b+c)b^(c+a)c^(a+b)
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