Prove $\sin x + \arcsin x > 2x$ using Maclaurin series My teacher asked us to solve this problem using the Maclaurin series, but I could not figure out how to approach..
Prove that the inequality sin x + arcsin x > 2x holds for all values of x such
that 0 < x ≤ 1.
I know that the Maclaurin series of 
sin(x) = x - $\frac{x^3}{3!}$ + $\frac{x^5}{5!}$ - $\frac{x^7}{7!}$ + ...
arcsin(x) = x + $\frac{1}{2}\cdot\frac{x^3}{3}$ + ($\frac{1}{2}\cdot\frac{3}{4}$)$\cdot\frac{x^5}{5}$ + ...
However, I do not know how to prove this using there series...Could anyone have some ideas?
Thank you!

For $0\le x\le1$ we have
$$\sin x\ge x-{1\over6}x^3\ge0\quad\text{and}\quad\arcsin x\ge x+{1\over6}x^3+{3\over40}x^5\ge0$$
which imply
$$\sin x\arcsin x\ge x^2+\left({3\over40}-{1\over36} \right)x^6-{1\over80}x^8=x^2+\left(34-9x^2\over720\right)x^6\ge x^2$$
By AGM we have
$${\sin x+\arcsin x\over2}\ge\sqrt{\sin x\arcsin x}\ge x$$
Remark: As Martin R astutely observes in comments, as soon as you have $\sin x\ge x-{1\over6}x^3$ and $\arcsin x\ge x+{1\over6}x^3$, you have $\sin x+\arcsin x\ge2x$, so tacking another (nonnegative) term onto the arcsine series, taking the product and using AGM is wholly unnecessary.  I failed to notice this because I was approaching things backwards: I had decided to see if AGM could be used and then worked out how much of the two series were needed to arrive at the desired inequality.
The inequality $\sin x\ge x-{1\over6}x^3$ for $0\le x\le1$ can be seen from the fact that the series for $\sin x$ is an alternating series of decreasing terms. 
