prove that $ x-\frac{1}{6}x^3<\sin(x)
prove that : 
  there exists a deleted neighborhood of $x=0$
  such that :
  $$
x-\frac{1}{6}x^3<\sin(x)<x-\frac{1}{6}x^3+\frac{1}{120}x^5
$$
MyTry:
let:$$f(x):=\sin x-x+\dfrac{1}{6}x^3$$
And :
$$g(x):=\sin x-x+\dfrac{1}{6}x^3-\dfrac{1}{125}x^5$$
Now what ?

From $$3^k\,\sin\frac{x}{3^k}-3^{k-1}\,\sin\frac{x}{3^{k-1}}=4\cdot3^{k-1}\,\sin^3\frac{x}{3^k},$$ by summing the telescoping series,
$$x-\sin x=4\,\sum^\infty_{k=1}3^{k-1}\,\sin^3\frac{x}{3^k}.$$ Since $\sin x<x$ for $x>0$, the RHS is $\le x^3/6$.
BTW, the inequality for $\sin x$ can be shown in a similar way, because
$$\frac{\sin x}{x}=\prod^\infty_{k=1}\cos\frac{x}{2^k}<1$$ for $x\neq0$.
