求证a^2+1>2aRT

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求证a^2+1>2aRT
求证a^2+1>2a
RT

求证a^2+1>2aRT
因为(a-1)²≥0
所以a²-2a+1≥0
所以a²+1≥2a
a=1时可以取等号

这个不能证明，只能证明a²+1≥2a
(a²+1)-2a
=a²-2a+1
=(a-1)²
≥0
∴a²+1≥2a

令x=a+1/a
则x²=a²+1/a²+2
所以即证明√(x²-2)-√2>=x-2
即证明√(x²-2)+2>=x+√2
即证明[√(x²-2)+2]²>=(x+√2)²
即证明(x²-2)+4√(x²-2)+4>=x²+2√2x+2
即...

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令x=a+1/a
则x²=a²+1/a²+2
所以即证明√(x²-2)-√2>=x-2
即证明√(x²-2)+2>=x+√2
即证明[√(x²-2)+2]²>=(x+√2)²
即证明(x²-2)+4√(x²-2)+4>=x²+2√2x+2
即证明4√(x²-2)>=2√2x
即证明2√(x²-2)>=√2x
即证明[2√(x²-2)]²>=(√2x)²
即证明4x²-8>=2x²
即证明x²>=4
因为a>0
所以x=a+1/a>=2√(a*1/a)=2
所以x²>=4成立
倒推回去
有√(a²+1/a²)-√2>=(a+1/a)-2
打字不易，如满意，望采纳。

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