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首先證明左半部分,應用放縮法: 1/√(1+a)+1/√(1+b)+1/√(1+c)>1/√(1+8)+1/√(1+8)+1/√(1+8)=1 再證明右半部分,還是應用放縮法: 1/√(1+a)+1/√(1+b)+1/√(1+c) 當其中的兩個趨近於0,另一個無窮大的時候,有最大值,最大值是2 所以整個式子
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