Compute $\lim_{n\to\infty} \int_{0}^{1}{\frac{n^{3/2}t}{1+n^3t^3}dt}$ using Dominated Convergence Theorem

Question

I'm having trouble with the dominating function.

The book suggests:

$$\frac{n^{3/2}t}{1+n^3t^3} \leq \frac{1}{2\sqrt{t}}$$

What should I do to prove this inequality?

Thanks in advance!

Answer 1

From $0\leq (a-b)^2$ it follows that $$ ab\leq \frac{1}{2}(a^2+b^2). $$ With $a=1$ and $b=n^{3/2}t^{3/2}$, you get $$ n^{3/2}t=\frac{1}{\sqrt{t}}(n^{3/2}t^{3/2})\leq \frac{1}{2\sqrt{t}}(1+n^3t^3). $$ Now divide by $1+n^3t^3$ and you are done.