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Use infinite series to prove that

$$\arcsin{x}\lt \frac{x}{1-x^2},$$ for $0\lt x\lt1$.

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    Try plugging in the infinite series expansion for arc sin. If this is not famous move from the expansion of sin. – 2017-02-20

3 Answers 3

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The first few terms of the Taylor series expansion of $\arcsin(x)$ are: $x + \frac{x^3}{6}+ \frac{3x^5}{40}+\frac{5x^7}{112}+\frac{35x^9}{1152}...$

The first few terms of the Taylor series expansion of $\frac{x}{1-x^2}$ are: $x + x^3 + x^5 + x^7 ...$

Obviously, the inequality is true.

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    Why all coefficients $\in(0,1)$? – 2017-02-20
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    Sorry, I do not understand what you are asking ? – 2017-02-20
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    @ Ekesh Why all coefficients like $\frac{1}{6}$, $\frac{3}{40}$, $\frac{5}{112}$,... in your Taylor series will be in $(0,1)?$ – 2017-02-20
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    The coefficients follow from the formula for Taylor series. – 2017-02-20
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    @Why all coefficients in your Taylor series will be in $[0,1)$? – 2017-02-20
1

From AM-GM inequality we have, for a given $t$ in the interval $(0,1)$,

$$\frac12\left[\frac{1}{(1+t)^2}+\frac{1}{(1-t)^2}\right]\ge \frac1{(1+t)(1-t)}\tag{1}$$

Now, since $1>\sqrt{(1+t)(1-t)}$, it follows $$\frac1{(1+t)(1-t)}>\frac{\sqrt{(1+t)(1-t)}}{(1+t)(1-t)}=\frac1{\sqrt{(1+t)(1-t)}}=\frac1{\sqrt{1-t^2}}\tag{2}$$

From $(1)$ and $(2)$ $$\frac12\left[\frac{1}{(1+t)^2}+\frac{1}{(1-t)^2}\right]>\frac1{\sqrt{1-t^2}}$$ Let $0\int_0^x\frac1{\sqrt{1-t^2}}dt\\ \frac12\left[-\frac{1}{1+x}+\frac1{1-x}+1-1\right]&>\arcsin x-0\\ \frac{x}{1-x^2}&>\arcsin x \end{align*}

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Let $f(x)=\frac{x}{1-x^2}-\arcsin{x}$.

Hence, $$f'(x)=\frac{1-x^2-x(-2x)}{(1-x^2)^2}-\frac{1}{\sqrt{1-x^2}}=\frac{1+x^2-\sqrt{(1-x^2)^3}}{(1-x^2)^2}=$$ $$=\frac{ x^2(x^4-2x^2+5)}{(1+x^2-\sqrt{(1-x^2)^3})(1-x^2)^2}>0.$$ Thus, $f(x)>f(0)=0$ and we are done!