<s>[INST] Prove $\sin x\ge x-\frac{x^3}{3!}$ 
Prove $$\sin x\ge x-\frac{x^3}{3!}$$ for $x\ge 0$.

I know that a calculus/continuity proof exists, but I am curious if this can be proved without that. Here is a sketch I have made.;
we will use for $x\ge 0$;

$$\sin x\le x\le \tan x$$

proof of this see here

I got a weaker bound :
$$\tan\left(\frac{x}{2}\right)\ge \frac{x}{2}$$
$$\sin\left(\frac{x}2\right)\cos\left(\frac{x}2\right)\ge \frac{x}{2}\cos^2\left(\frac{x}{2}\right)$$
$$\sin x= x\left[1-\sin^2\left(\frac{x}{2}\right)\right]\ge x\left(1-\frac{x^2}{4}\right)= x-\frac{x^3}{4}$$
Thus $$\sin x \ge x-\frac{x^3}{4}$$
As you can see is there any way I can take this to strengthen the inequality, to get $$\sin x\ge x-\frac{x^3}{6}.$$
Also could this possibly be extended by strengthening inequality further to get the Taylor series. I think I am going too far!
 [/INST] Well, from the celebrated triplication formula, we get $$\sin3x=3\,\sin x-4\,\sin^3x.$$ This, of course, means that $$3^{k}\sin x/3^k-3^{k+1}\sin x/3^{k+1}=-4\cdot3^{k}\,\sin^3 x/3^{k+1}.$$ Summing the telescopic series and observing $3^{k}\sin x/3^k\to x$ as $k\to\infty$, we get
$$\sin x-x= -4\,\sum^\infty_{k=0}3^{k}\,\sin^3 x/3^{k+1},$$ i.e.
$$x-\sin x= 4\,\sum^\infty_{k=0}3^{k}\,\sin^3 x/3^{k+1}.$$
Since $\sin x\le x$ for non-negative $x$, this means
$$x-\sin x\le4\,\sum^\infty_{k=0}3^k\,\left(x/3^{k+1}\right)^3=4\cdot\frac1{27}\cdot\frac1{1-\frac19}\,x^3=\frac16\,x^3.$$
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