欲证√[a^2+1/(a^2)]-√2>a+1/a-2,则证{√[a^2+1/(a^2)]-√2}/(a+1/a-2)>1;
分子有理化,得:{(a^2+1/a^2-2)}/{(a+1/a-2)(√[a^2+1/(a^2)]+√2)}
=>{(a-1/a)^2}/{(a+1/a-2)(√[a^2+1/(a^2)]+√2)}
=>{(√a+1/√a)^2*(√a-1/√a)^2}/{(a+1/a-2)(√[a^2+1/(a^2)]+√2)}
=>{(√a+1/√a)^2}/{(√[a^2+1/(a^2)]+√2)}
=>{(a+1/a+2)}/{(√[a^2+1/(a^2)]+√2)}
∵(a+1/a)^2≥a^2+1/(a^2)
∴(a+1/a)≥√[a^2+1/(a^2)]
又2>√2所以原式得证
大致就是这个步骤,这样看看不明白的,你写在纸上看,那个除号是长分号,有什么问题再问
打了半天,