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. 
 A: Note that $$f(n)-g(n)=\frac{2}{3}(n^{3/2}-1)+\sqrt{n}-1-\frac{2}{3}((n+1)^{3/2}-2^{3/2})=\int_1^n\left(\sqrt{x}+\frac1{2\sqrt{x}}-\sqrt{x+1}\right)\mathrm dx,$$ hence it suffices to show that, for every $x\geqslant1$, $$\sqrt{x}+\frac1{2\sqrt{x}}-\sqrt{x+1}\geqslant0.$$ The LHS is $$\frac1{2\sqrt{x}}-(\sqrt{x+1}-\sqrt{x})=\frac1{2\sqrt{x}}-\frac1{\sqrt{x+1}+\sqrt{x}}=\frac{\sqrt{x+1}-\sqrt{x}}{2\sqrt{x}(\sqrt{x+1}+\sqrt{x})}\gt0,$$ hence the proof is complete.
A: set $h(x)=f(x)-g(x)=\Big( \frac{2}{3}(x^{3/2}-1)+\sqrt{x} \Big)- \Big(1+\frac{2}{3}((x+1)^{3/2}-2^{3/2})\Big).$
Differentiate w.r.t x, we get $h'(x)=\frac{x+1/2-\sqrt{x(x+1)}}{\sqrt{x}}>0$
So $h$ is strictly increasing function. Then check $h(1)=0$, and we get the result.
A: 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:
$$ 1-\frac{1}{8(n+1)^2}+\frac{1}{(8n+14)(n+1)\left(1-\frac{1}{3(n+1)^2}\right)}\leq 1.$$
A: Let $h(x)=\sqrt{x}$.  
Then $\displaystyle f(n)\ge g(n)\iff\int_{1}^{n}h(x)dx+\sqrt{n}\ge1+\int_{2}^{n+1}h(x)dx\iff$
$\displaystyle{\hspace 1.4 in}\int_{1}^{2}h(x)dx-h(1)\ge\int_{n}^{n+1}h(x)dx-h(n)$,
so let $\;\;\displaystyle k(x)=\int_{x}^{x+1}h(t)dt-h(x)$.
Since $k^{\prime}(x)=h(x+1)-h(x)-h^{\prime}(x)=h^{\prime}(c)-h^{\prime}(x)$ for some $c$ with $x<c<x+1$
and $h$ is concave down, $k^{\prime}(x)<0$ and so $k$ is decreasing. 
Therefore $k(1)\ge k(n)$ for all $n\ge1$, so $f(n)\ge g(n)$ for all $n\ge1$.
