Part of the exercise from "Algorithms Illuminated" book.
Arrange the following functions in order of increasing growth rate, with g(n) following f (n) in your list if and only if f (n) = O(g(n)).
a) $2^\sqrt{log_2n}$
b) $n^{\frac{5}{3}}$
We can suggest, that $2^\sqrt{log_2n}$ = O ( $n^{\frac{5}{3}}$ ), then
$$2^\sqrt{log_2n} \leqslant c \cdot n^{\frac{5}{3}}$$Take $log_2n$ of both sides
$$log_2(2^\sqrt{log_2n}) \leqslant log_2(c) \cdot log_2(n^{\frac{5}{3}})$$
$$\sqrt{log_2n} \leqslant log_2(c) \cdot \frac{5}{3}log_2(n)$$Square both sides, and get rid of 5/3, since it's big-O, and constant can be neglected
$$log_2n \leqslant (log_2(c))^2 + 2log_2(c)log_2(n) + (log_2(n))^2$$
Now we can divide both sides by $log_2(n)$
$$1 \leqslant \frac{(log_2(c))^2}{log_2n} + 2log_2c + log_2 (n)$$
Is it a valid proof to show, that a) is a Big-O of b) Or it's a total gibberish, that doesn't prove anything, and I've messed it all up?