求证(a+b/2)²≤a²+b平方/2
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求证(a+b/2)²≤a²+b平方/2
求证(a+b/2)²≤a²+b平方/2
求证(a+b/2)²≤a²+b平方/2
作差,得:
(1/4)(a+b)²-(1/2)(a²+b²)
=(1/4)[(a+b)²-2(a²+b²)]
=(1/4)[-a²+2ab-b²]
=-(1/4)(a-b)²≤0
即:
(1/4)(a+b)²≤(1/2)(a²+b²)