Prove that arcsin x > x+ (x³)/6 for x in (0,1)

Prove that $\arcsin x\gt x+\dfrac{x^{3}}{6}$ for $x\in\bigl(0,1\bigr)$.

Step 1: The auxiliary function

\[ f(x) = \arcsin x-x-\frac{x^{3}}{6}, \qquad x\in\bigl[0,1\bigr), \qquad f(0) = 0. \]

It suffices to show that $f$ is strictly increasing on $\bigl(0,1\bigr)$.

Step 2: The first derivative, which needs its own argument

\[ g(x) = f'(x) = \frac{1}{\sqrt{1-x^{2}}}-1-\frac{x^{2}}{2}, \qquad g(0) = 0. \]

Its sign is not obvious, so we differentiate once more.

Step 3: The second derivative is clearly positive

\[ g'(x) = \frac{x}{\bigl(1-x^{2}\bigr)^{3/2}}-x = x\left[\frac{1}{\bigl(1-x^{2}\bigr)^{3/2}}-1\right]. \]

For $0\lt x\lt 1$ we have $0\lt 1-x^{2}\lt 1$, hence $\bigl(1-x^{2}\bigr)^{3/2}\lt 1$ and the bracket is positive; and $x\gt 0$. Therefore $g'(x)\gt 0$ on $\bigl(0,1\bigr)$.

Step 4: Climb back up the chain

$g$ is strictly increasing with $g(0) = 0$, so $g(x)\gt 0$ on $\bigl(0,1\bigr)$; that is, $f'\gt 0$ there. Then $f$ is strictly increasing with $f(0) = 0$, so $f(x)\gt 0$ on $\bigl(0,1\bigr)$, which is the assertion.

Step 5: Verification

\[x = 0.5:\ 0.5235988 \gt 0.5208333; \qquad x = 0.9:\ 1.1197695 \gt 1.0215 \;\;\checkmark\]

\[x = 0.99:\ 1.4292569 \gt 1.1517165 \;\;\checkmark\ \text{(the gap widens sharply near } 1)\]

\[\text{Series: } \arcsin x = x+\frac{x^{3}}{6}+\frac{3x^{5}}{40}+\cdots,\ \text{all coefficients positive} \;\;\checkmark\]