Is it true that $$\sin x > \dfrac x{\sqrt {x^2+1}} , \forall x \in \left(0, \dfrac {\pi}2\right)$$ (I tried differentiating , but it's not coming , please help)


$\arcsin$ is an increasing function, and you can rewrite


Then, deriving (and using equality at $x=0$)


The original inequation is a mixture of trigonometric and algebraic expressions, which makes it intractable as such. The "trick" is to let the sine disappear, knowing that the derivative of the $\arcsin$ is an algebraic expression.

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    $\begingroup$ Interestingly, this is a way to "discover" that $\arcsin(x/\sqrt{x^2+1})=\arctan(x)$. $\endgroup$ – Yves Daoust Jul 5 '15 at 9:14

How about the substitution $x = \tan \theta$, then the inequality reduces to proving:

$$\sin \tan \theta \ge \sin \theta$$

for, $\theta \in \left(0,\tan^{-1} (\pi/2\right))$.

Since, $\sin \theta$ is monotone increasing in the interval $\left(0,\pi/2\right)$, we are done using: $$\tan \theta \ge \theta$$

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    $\begingroup$ You probably mean $\theta\in(0,\arctan(\pi/2))$. $\endgroup$ – Yves Daoust Jul 5 '15 at 9:28
  • $\begingroup$ @YvesDaoust $tan^{-1}\pi/2 \le \pi/2$, so the first inequality holds true in that interval too I supposed. $\endgroup$ – r9m Jul 5 '15 at 14:58
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    $\begingroup$ No, $1.4<\pi/2$, but $\sin(\tan(1.4))<\sin(1.4)$. $\endgroup$ – Yves Daoust Jul 6 '15 at 6:32
  • $\begingroup$ @YvesDaoust ah! right you are!! Thanks!! $\endgroup$ – r9m Jul 6 '15 at 6:35

Square both sides (assume $x\in(0,\pi/2)$):

$$\sin x>\frac{x}{\sqrt{x^2+1}}\iff 1-\cos^2 x>1-\frac{1}{x^2+1}$$

$$\iff \cos ^2 x<\frac{1}{x^2+1}=\frac{\sin^2 x+\cos^2 x}{x^2+1}$$

$$\iff x^2\cos^2 x+\cos^2 x<\sin^2 x + \cos^2 x$$

$$\iff \tan^2 x> x^2\iff \tan x>x,$$

which is true ($(\tan x-x)'>0$ for $x\in(0,\pi/2)$ and $\tan(0)-0=0$).

  • $\begingroup$ You can straigthen up the derivation by noting that $(\cos^2x)^{-1}=1+\tan^2x$, so the fourth step is directly $1+\tan^2x>x^2+1$. $\endgroup$ – egreg Jul 6 '15 at 7:15

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