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!

Hint: A nicer representation of the arcsin series is 
$$\arcsin(x) = \sum_{n = 0}^\infty \frac{1}{4^n} {2n \choose n} \cdot \frac{x^{2n + 1}}{2n + 1}$$
Since both $\sin(x)$ and $\arcsin(x)$ are both odd, we can look at the coefficients on all the odd-degree terms. Can you prove that for all $n \geq 1$, 
$$\frac{1}{4^n} {2n \choose n} \cdot \frac{1}{2n + 1} \geq \frac{1}{(2n + 1)!} \iff \frac{1}{4^n} {2n \choose n} \geq \frac{1}{(2n)!}$$ 
This would prove all coefficients of terms with degree greater than $1$ in the expansion of $\arcsin(x) + \sin(x)$ are positive, and thus $\sin(x) + \arcsin(x) = 2x + C$, where $C$ is strictly positive.
