(本小题12分)证明:要证:2(a2+b2)≥(a+b)2
只要证2a2+2b2≥a2+b2+2ab
只要证a2+b2≥2ab (5分)
即(a-b)2≥0,而此式显然成立
所以2(a2+b2)≥(a+b)2成立(12分)
已知a,b∈R,求证2(a2+b2)≥(a+b)2.
已知a,b∈R,求证2(a2+b2)≥(a+b)2.
数学人气:305 ℃时间:2020-02-05 14:01:29
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