Proving $\displaystyle \frac{\sin^3x}{x}\lt 0.69$ for any $x\gt 0$

Question

Question : How can we prove strictly that the following inequality holds for any $x\gt0$?$$\frac{\sin^3x}{x}\lt 0.69$$

This seems difficult though it doesn't look so. Can anyone help?

Answer 1

Hint

Since $\sin(x)$ is bounded and that $x$ is permanently increasing, you basically look for the first maximum of $$f(x)=\frac{\sin^3x}{x}$$ The derivative is given by $$f'(x)=\frac{\sin ^2(x) (3 x \cos (x)-\sin (x))}{x^2}$$ which cancels at the place where $\tan(x)=3x$; the root of this equation corresponds to the intersection of the curve $y=\tan(x)$ and the straight line $y=3x$. If you plot both functions, you should notice that the solution, which must be between $0$ and $\frac{\pi}{2}$, is close to $x=1.3$. If you polish the root using Newton method for example, you will find that the derivative cancels for $x=1.32419$ and, for this value, $f(x)=0.688691$. The second derivative test would confirm that this is a maximum of the function.

Answer 2

To simplify the notations, we introduce the function $$ f(x) = \frac{\sin^{3}(x)}{x} $$ We have the usual inequality $$ |\sin(x)| \leq 1 $$ so that for $|x| \geq \frac{1}{0.68}$, we have $|f(x)| < 0.68$. As a consequence, for $|x| \geq \frac{1}{0.68}$, the inequality is satisfied.

The second remark is that $x \mapsto f(x)$ is an even function, so that it only remains to study the behavior of this function for $x \in [0 \,,\, \frac{1}{0.68}]$.

The derivative of $f$ is straightforwardly given by $$ f'(x) = \frac{(3 x \cos(x) - \sin(x)) \sin(x)^{2}}{x^{2}} $$

The maximum of $f$ is reached for $x_{\rm max}$ which solves $$ 3 x_{\rm max} \cos(x_{\rm max}) - \sin(x_{\rm max}) = 0$$

At this stage, numerical estimations give you that $$ x_{\rm max} \simeq 1.32419 $$ so that $$ f (x_{\rm max}) \simeq 0.688691 < 0.69 $$ (This check-up is made numerically, but it can be made 100% bullet-proof !)

So that finally $\forall x \in \mathbb{R}, |f (x) | < 0.69$, which ensures immediately $f(x) < 0.69$