I have the following inequality that I am trying to prove:
$log_2n < 2*(log_2n)^{1/2}$
${log_2n \over 2*(log_2n)^{1/2}} < 1$
I am trying to find a value N1 > 0 so that the inequality is true for all n > N1.
My approach is to use L'Hopital's rule so that if I can show that $\lim_{x\to ∞} f'(n)/g'(n) = 0$ then the inequality will hold.
$\lim_{x\to ∞} f(n)/g(n) = ∞/∞$, so L'Hopital's rule can be used.
However, using L'Hopital's rule I have got:
$\lim_{x\to ∞} f'(n)/g'(n) = {{1 \over nlog2}/{1\over n*(log2)^{1/2}*(logn)^{1/2}}} = {1/∞ \over 1/∞}$ = 0/0
which is of indeterminate form.
What are some other ways that I can prove the inequality?
The inequality is not true. In fact, $\log_2(x) > 2(\log_2(x))^\frac{1}{2}$ for all $x > 16$.