prove logarithmic inequality for N>1200

Question

For N > 1200 how can i prove that

3.09N/Log(N) + 1 <= 1.7(2N+1)/Log(2N+1)

(sorry, could not figure out how to put the 'less than or equal' symbol there, tried \leq)

Answer 1

Just by derivation: let $$f(x):=1.7\frac{2x+1}{\log (2x+1)} - 3.09\frac{x}{\log x}-1,$$ $$f'(x)= 3.4 \frac{\log(2x+1)-1}{\log^2(2x+1)} - 3.09 \frac{\log x-1}{\log^2x}. $$

1. need to check $f'(x)>0$ for $x>1200$, hence $$3.4 \frac{\log(2x+1)-1}{\log^2(2x+1)} - 3.09 \frac{\log x-1}{\log^2x} \geq 0$$ $$\frac{3.4}{3.09}\geq \frac{\log x-1}{\log^2x} \frac{\log^2(2x+1)}{\log(2x+1)-1} \geq \frac{\log x-1}{\log^2x} \frac{\log^2(2x+1)}{\log(2x+1)}=\frac{\log x-1}{\log^2x}\log (2x)$$ $$ = \frac{\log 2+\log x}{\log x}-\frac{\log (2x)}{\log^2x}= 1+\frac{\log 2}{\log x}-\frac{\log (2x)}{\log^2x}.$$ Since $x\mapsto \frac{\log 2}{\log x}$ is decreasing, we get $\frac{\log 2}{\log x} \leq \frac{\log2}{\log1200} \approx 0.097763$. Since $\frac{3.4}{3.09}>1+0.097763$, we have $f'(x)>0$ for $x\geq 1200$.

2. since $f(1200)=1.7\frac{2401}{\log 2401} - 3.09\frac{1200}{\log 1200}-1>0$, your inequality follows.