题目(2001年全国卷Ⅱ,理20)已知m,n∈N+,且mn.证明:(1+m)n(1+n)m.试题初看平淡无奇,深入探讨后发现试题解法众多,方法灵活并且内涵丰富.命题者提供的参考解答过程显得有些繁琐.笔者提供数形结合的巧证,供参考.巧证原不等式等价于nln(1+m)mln(1+n),即ln(1m+m)ln(1n+n)(1≤m The subject (2001 National Volume II, Li 20) known m, n∈N +, and mn. Proof: (1 + m) n (1 + n) m. Many, method is flexible and connotation is rich. The proposition person provides the reference solution process somewhat complicated. The author provides the digit combination form the clever evidence, for the reference.Qualifying the original inequality is equivalent to nln (1 + m) mln (1 + n) , That is, ln (1m + m) ln (1n + n) (1≤m