∵Z=[4x^2+(4m+4n-1)x-(5m+2n-1)] / [(4m+4n+1)x-(5m+2n-1)]=1+(4x^2-2x)/ [(4m+4n+1)x-(5m+2n-1)]＞1
∴F(x)={[4x^2+(4m+4n-1)x-(5m+2n-1)] / [(4m+4n+1)x-(5m+2n-1)]} ^(m/x)＜1+(m/x)×
(4x^2-2x)/ [(4m+4n+1)x-(5m+2n-1)]=1+(4mx-2m)/ [(4m+4n+1)x-(5m+2n-1)
又∵A=(4mx-2m)/ [(4m+4n+1)x-(5m+2n-1)是減函數(shù)(可用導(dǎo)數(shù)證)
∴F(x)＜1+(4m×2-2m)/ [(4m+4n+1)×2-(5m+2n-1)=1+6m/(3m+6n+3)＜1+[m/(n+1)]
不好意思,這部分F(x)={[4x^2+(4m+4n-1)x-(5m+2n-1)] / [(4m+4n+1)x-(5m+2n-1)]} ^(m/x)＜1+(m/x)×(4x^2-2x)/ [(4m+4n+1)x-(5m+2n-1)]不太會(huì),是不是應(yīng)該用導(dǎo)數(shù)證?
根據(jù)樓主的提示,證{1+[2/(m+2n+1)]}^m < 1+[m/(n+1)]
設(shè):y={1+[2/(m+2n+1)]}^m ,y'=m{1+[2/(m+2n+1)]}^(m-1)×(-2)+{1+[2/(m+2n+1)]}^m×ln{1+[2/(m+2n+1)]}={1+[2/(m+2n+1)]}^(m-1)×【{1+[2/(m+2n+1]}ln{1+[2/(m+2n+1)]}-2m】
當(dāng)m、n∈N+時(shí),y‘＜0,y為減函數(shù),當(dāng)m=1時(shí),y最大
∴y≤1+2/（1+2n+1）=1+[1/(n+1)]＜1+[m/(n+1)]