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与自然数n有关的不等式的证明通常采用数学归纳法。这里我们给出可与数学归纳法相媲美的新方法——自然数函数单调性法。定理若n、n_0∈N,且n>n_0,f(n)是自然数n的单调递增(或单调递减)函数且f(n_0)≥m(或≤M),则f(n)≥m(或≤M)。由函数的单调性知上面的定理是显然的,下面举例说明它的应用。例1 求证:当n是不小于3的整数时,有n~(n+1)>(n+1)~n。证明设f(n)=((n+1)~n)/(n~(n+1)),
The proof of the inequality associated with the natural number n usually uses mathematical induction. Here we present a new method that can be compared with mathematical induction—the monotonicity of natural number functions. Theorem if n, n_0∈N, and n>n_0, f(n) is a monotonically increasing (or monotonically decreasing) function of natural number n and f(n_0)≥m(or ≤M), then f(n)≥m( Or ≤ M). It is obvious that the above theorem is known by the monotonicity of the function, and its application is illustrated below. Example 1 Verification: When n is an integer not less than 3, there are n~(n+1)>(n+1)~n. Proof Let f(n)=((n+1)~n)/(n~(n+1)),