已知实数a,b,c满足a+b+c=1,求证:(根号3a=1)+(根号3b+2)+(根号3c+3)≤3(根3) 已知实数a,b,c满足a+b+c=1,求证:(根号3a+1)+(根号3b+2)+(根号3c+3)≤3(根3)
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![已知实数a,b,c满足a+b+c=1,求证:(根号3a=1)+(根号3b+2)+(根号3c+3)≤3(根3) 已知实数a,b,c满足a+b+c=1,求证:(根号3a+1)+(根号3b+2)+(根号3c+3)≤3(根3)](/uploads/image/z/1012334-14-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9E%E6%95%B0a%2Cb%2Cc%E6%BB%A1%E8%B6%B3a%2Bb%2Bc%3D1%2C%E6%B1%82%E8%AF%81%3A%28%E6%A0%B9%E5%8F%B73a%3D1%29%2B%28%E6%A0%B9%E5%8F%B73b%2B2%29%2B%28%E6%A0%B9%E5%8F%B73c%2B3%29%E2%89%A43%28%E6%A0%B93%29+%E5%B7%B2%E7%9F%A5%E5%AE%9E%E6%95%B0a%2Cb%2Cc%E6%BB%A1%E8%B6%B3a%2Bb%2Bc%3D1%2C%E6%B1%82%E8%AF%81%3A%28%E6%A0%B9%E5%8F%B73a%2B1%29%2B%28%E6%A0%B9%E5%8F%B73b%2B2%29%2B%28%E6%A0%B9%E5%8F%B73c%2B3%29%E2%89%A43%28%E6%A0%B93%29)
已知实数a,b,c满足a+b+c=1,求证:(根号3a=1)+(根号3b+2)+(根号3c+3)≤3(根3) 已知实数a,b,c满足a+b+c=1,求证:(根号3a+1)+(根号3b+2)+(根号3c+3)≤3(根3)
已知实数a,b,c满足a+b+c=1,求证:(根号3a=1)+(根号3b+2)+(根号3c+3)≤3(根3)
已知实数a,b,c满足a+b+c=1,求证:(根号3a+1)+(根号3b+2)+(根号3c+3)≤3(根3)
已知实数a,b,c满足a+b+c=1,求证:(根号3a=1)+(根号3b+2)+(根号3c+3)≤3(根3) 已知实数a,b,c满足a+b+c=1,求证:(根号3a+1)+(根号3b+2)+(根号3c+3)≤3(根3)
这道题可以用逆向思维,反证法:
要证明(根号3a+1)+(根号3b+2)+(根号3c+3)≤3(根3),即要证明:
[(根号3a+1)+(根号3b+2)+(根号3c+3)]^2≤[3(根3)]^2
展开得:(3a+1)+(3b+2)+(3c+3)+2[根号(3a+1)(3b+2)]+2[根号(3a+1)(3c+3)]+2[根号(3b+2)(3c+3)]≤27;
而:(3a+1)+(3b+2)+(3c+3)=3(a+b+c)+6=9;
再由均值不等式得:(3a+1)+(3b+2)>=2[根号(3a+1)(3b+2)];
(3a+1)+(3c+3)>=2[根号(3a+1)(3c+3)];
(3b+2)+(3c+3)>=2[根号(3b+2)(3c+3)];
这三个不等式在3a+1=3b+2=3c+3时,取等号!
将这三个不等式相加:6(a+b+c)+16>=2[根号(3a+1)(3b+2)]+2[根号(3a+1)(3c+3)]+2[根号(3b+2)(3c+3)]
即:18>=2[根号(3a+1)(3b+2)]+2[根号(3a+1)(3c+3)]+2[根号(3b+2)(3c+3)]
综合上述,得到:(3a+1)+(3b+2)+(3c+3)+2[根号(3a+1)(3b+2)]+2[根号(3a+1)(3c+3)]+2[根号(3b+2)(3c+3)]≤27
两边再开根号,原命题得证!