题目内容
已知x∈R,求证:1+2x4≥x2+2x3.
证法一:1+2x4-(x2+2x3)
=2x3(x-1)-(x-1)(x+1)
=(x-1)(2x3-x-1)
=(x-1)2(2x2+2x+1)
=(x-1)2［(x+1)2+x2］≥0,
即1+2x4-(x2+2x3)≥0.
所以1+2x4≥x2+2x3.
证法二:1+2x4-(x2+2x3)
=x4-2x3+x2+x4-2x2+1
=x2(x-1)2+(x2-1)2≥0.
即1+2x4-(x2+2x3)≥0.
所以1+2x4≥x2+2x3.
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