用分析法证明:若a>0,则根号下(a^2+1/a^2)-根号2>a+1/a-2

2个回答

  • 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,