How can I prove this inequality by M.I. or otherwise? This question is from the past examination:
$$f(n)=\frac{2}{3}(n^{3/2}-1)+\sqrt{n}$$
$$g(n)=1+\frac{2}{3}((n+1)^{3/2}-2^{3/2})$$
My task to prove $f(n)≥g(n)$ for all $n≥1$.
I have tried M.I here.
But the problem is that I do not know how to expand the term $(n+2)^{3/2}$ in the inductive steps. Can anyone tell me what method should I use to finish the proof. 

Since $f(n)\geq g(n)$ holds for $n=1$, it is sufficient to show that:
$$\begin{eqnarray*} f(n+1)-f(n) &=& \frac{2}{3}((n+1)^{3/2}-n^{3/2})+(n+1)^{1/2}-n^{1/2}\\&\geq& \frac{2}{3}((n+2)^{3/2}-(n+1)^{3/2})=g(n+1)-g(n)\end{eqnarray*}$$
that is equivalent to:
$$\frac{2}{3}\left((n+2)^{3/2}-2(n+1)^{3/2}+n^{3/2}\right)\leq (n+1)^{1/2}-n^{1/2}\tag{1}$$
or to:
$$(2n+4)(n+2)^{1/2}+(2n+3)n^{1/2}\leq (4n+7)(n+1)^{1/2}\tag{2}$$
that is equivalent to:
$$\frac{2n+4}{4n+7}\left(1+\frac{1}{n+1}\right)^{1/2}+\frac{2n+3}{4n+7}\left(1-\frac{1}{n+1}\right)^{1/2}\leq 1. \tag{3}$$
In virtue of the Cauchy-Schwarz inequality the LHS of $(3)$ is less or equal than:
$$\frac{\sqrt{2}}{4n+7}\cdot\sqrt{(2n+4)^2+(2n+3)^2}=\sqrt{1+\frac{1}{(4n+7)^2}},$$
that, however, is bigger than one. Using $\sqrt{1+x}+\sqrt{1-x}\leq 2-\frac{x^2}{4}$ we get that the LHS of $(3)$ is less or equal than:
$$1-\frac{1}{8(n+1)^2}+\frac{1}{8n+14}\left(\left(1+\frac{1}{n+1}\right)^{1/2}-\left(1-\frac{1}{n+1}\right)^{1/2}\right)$$
and since over $(0,1]$ we have $\sqrt{1+x}-\sqrt{1-x}\leq\frac{x}{1-x^2/3}$, it follows that the LHS of $(3)$ is less than:
$$ 1-\frac{1}{8(n+1)^2}+\frac{1}{(8n+14)(n+1)\left(1-\frac{1}{3(n+1)^2}\right)}\leq 1.$$
