Prove that $\sqrt{n+1}>\sqrt{n}+\frac{1}{2\sqrt{n}}-\frac{1}{8n\sqrt{n}}$ 
Prove that $\sqrt{n+1}>\sqrt{n}+\frac{1}{2\sqrt{n}}-\frac{1}{8n\sqrt{n}}$ if $n>0$. 

I didn't see an easy way of proving this without doing a lot of algebra and rearranging. Is there an easier way?
 A: It is hardly a lot of algebra. To save on MathJax, I write $\sqrt n$ as $s$. So the square of the RHS is $(s+\frac{1}{2s}-\frac{1}{8s^3})^2=1+s^2+\frac{1}{64s^6}-\frac{1}{8s^4}$. The square of the LHS is just $1+s^2$.
So you have to show that $\frac{1}{8s^4}>\frac{1}{64s^6}$ or $8s^2>1$ or $8n>1$ which is true for any positive integer and indeed any real $>\frac{1}{8}$.
A: It does not look so terrible to me! We have:
$$ \sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}} \tag{1}$$
and:
$$ \frac{1}{2\sqrt{n}}-\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n}\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{1}{2\sqrt{n}\left(\sqrt{n}+\sqrt{n+1}\right)^2}\tag{2}$$
so the inequality boils down to $\sqrt{n}+\sqrt{n+1}>2\sqrt{n}$, quite trivial.
A: For $y>0, (1+y)^{-3/2}<1$. Integrate both sides from $0$ to $x$:
$$2\left(1-(1+x)^{-1/2}\right)=\int_0^x(1+y)^{-3/2}\,dy<\int_0^x\,dy=x$$
which, upon rearranging, gives that $1-\frac{x}{2}<(1+x)^{-1/2}$. We now integrate from $0$ to $1/n$:
$$\frac{1}{n}-\frac{1}{4n^2}=\int_0^{1/n}\left(1-\frac{x}{2}\right)\,dx<\int_0^{1/n}(1+x)^{-1/2}\,dx=2\left(\left(1+\frac{1}{n}\right)^{1/2}-1\right)$$
which, upon further rearranging, gives:
$$\sqrt\frac{n+1}{n}=\left(1+\frac{1}{n}\right)^{1/2}>1+\frac{1}{2n}-\frac{1}{8n^2}$$
Multiplying by $\sqrt n$ gives the desired result.
