Solving explicitly or by induction: $0 \leq 2^{(n/2+10)}-2\pi n^{(3/2)}$ Is it possible to solve the following equation explicitly?
$$0 \leq 2^{(n/2+10)}-2\pi n^{(3/2)}$$
I know this is correct and the answer is $n \geq 0$.
I've also tried induction on $2\pi n^{(3/2)} \leq 2^{(n/2+10)}$, but with no success.
There is another method which involves taking successive derivatives and showing it's increasing but those equations themselves seem even harder to solve.
 A: We have
$$0 \leq 2^{(n/2+10)}-2\pi n^{(3/2)}\iff 2^{(n/2+10)}\ge 2\pi n^{(3/2)}\iff \log(2^{(n/2+10)})\ge\log(2\pi n^{(3/2)})$$
$$\iff \frac n 2 \log 2 +10\log 2 \ge \log 2+\log \pi+\frac 32\log n\iff\log n\le\frac n 3\log 2+6\log 2-\frac23\log \pi$$
which is true eventually and by inspection we see notably that it is true for $n=1,2,3,4$


*

*$n=1 \implies \log 1=0\le\frac 1 3\log 2+6\log 2-\frac23\log \pi\approx 3.63$

*$n=2 \implies \log 2\approx 0.69\le\frac 2 3\log 2+6\log 2-\frac23\log \pi\approx 3.85$

*$n=3 \implies \log 3\approx 1.10\le\frac 3 3\log 2+6\log 2-\frac23\log \pi\approx 4.10$

*$n=4 \implies \log 4\approx 1.39\le\frac 4 3\log 2+6\log 2-\frac23\log \pi\approx 4.32$
then consider
$$f(x)=\log x-\frac x 3 \log 2 - 6 \log 2 +\frac23 \log \pi\implies f'(x)=\frac 1x-\frac13\log 2<0 \quad x>\frac 3{\log 2}\approx 4.33$$
therefore the given inequality is true for any $n$.
A: $$2\pi n^{(3/2)} \leq 2^{(n/2+10)}$$
$$\log_2 (\pi n^{(3/2)}) \leq (n/2+9)$$
$$\log_2 (\pi)+\dfrac{3\log_2(n)}{2} \leq (n/2+9)$$
$$\log_2 (\pi)-9 \leq \dfrac{n- 3\log_2(n)}{2}$$
$$n- 3\log_2(n) \geq -14.69$$
$$3\log_2(n) - n \leq 14.69$$
Let $y=3\log_2(n)-n$
To see what is the maximum value of the left side of the inequality:
$$\dfrac{dy}{dn} = \dfrac{3}{n\ln(2)}-1 = 0$$
$$n = \dfrac{3}{ln(2)},y=2.01$$
To check if its the maximum value of y,
$$\dfrac{d^2y}{dn} = -\dfrac{ln(8)}{n^2} = -\dfrac{ln(8)}{(\dfrac{3}{\ln(2)})^2} = -0.11 < 0$$
As $max(y) \leq 14.69$ the inequality holds for all values of n.
