Mathematical induction - - - challenge problem

[lookagain]

Using the Principle of Mathematical Induction, prove \(\displaystyle \ (\frac{3}{4})\sqrt{n} \ > \ \ln(n), \ \ for \ \ all \ \ positive \ \ integers \ \ n.\)

 

[daon2]

Since \(\displaystyle f(n):=\dfrac{3}{4}\sqrt{n} > \ln(n):=g(n)\) for \(\displaystyle n=1,2,3\) and \(\displaystyle 4x\le x^2\) for \(\displaystyle x\ge4\), we get

\(\displaystyle f'(x)=\dfrac{3}{8\sqrt{x}}\ge \dfrac{1}{\sqrt{2x}} \ge \dfrac{1}{x}=g'(x)\).

 

[lookagain]

Since \(\displaystyle f(n):=\dfrac{3}{4}\sqrt{n} > \ln(n):=g(n)\) for \(\displaystyle n=1,2,3\) and \(\displaystyle 4x\le x^2\) for \(\displaystyle x\ge4\), we get

\(\displaystyle f'(x)=\dfrac{3}{8\sqrt{x}}\ge \dfrac{1}{\sqrt{2x}} \ge \dfrac{1}{x}=g'(x)\).

No, your solution is not acceptable. You must use induction, no calculus. Denis already gave the base step. What's missing are the last three steps: ii) assume it is true for n (or n = k), (iii) show it is true for n + 1 (or n = k + 1), and (iv) state the conclusion.