Show that $\frac{x}{3!}-\frac{x^3}{5!}+\frac{x^5}{7!}-\cdots\leq \frac{1}{\pi}$.

Solution 1:

Using $f(x)=\frac{x-\sin(x)}{x^2}$, the inequality is equivalent to: $$ \frac{x-\sin(x)}{x^2}≤\frac{1}{\pi}\iff x\left(1-\frac{x}{\pi}\right)≤\sin(x) $$ Thus, define $g(x):=x\left(1-\frac{x}{\pi}\right)$.

Note that $g(x)=g(\pi-x)$ and $\sin(x)=\sin(\pi-x)$. Therefore it suffices (by symmetry) to prove the inequality for $x\in\left(-\infty;\frac{\pi}{2}\right]$.

Firstly, we prove it for $x\in\left[0;\frac{\pi}{2}\right]$:

We have $g(0)=\sin(0)=0$, thus it is sufficient to prove, that $g'(x)≤\sin'(x)$ for $x\in\left[0;\frac{\pi}{2}\right]$. Now: $g'(x)=1-\frac{2x}{\pi}$ and $\sin'(x)=\cos(x)$. So it remains to prove: $$ 1-\frac{2x}{\pi}≤\cos(x) $$ Since $\cos''(x)=-\cos(x)≤0\space\forall x\in\left[0;\frac{\pi}{2}\right]$, $\cos(x)$ is concave on the interval. As $g'(0)=\cos(0)=1$ and $g'\left(\frac{\pi}{2}\right)=\cos\left(\frac{\pi}{2}\right)=0$, $g'(x)$ connects two points on the graph of $\cos(x)$, and together with the concavity, this implies the above inequality.

Secondly, we prove it for $x\in\left(-\infty;0\right]$:

By the same argument as above, we just have to show, that on this interval $g'(x)≥\sin'(x)\iff 1-\frac{2x}{\pi}≥\cos(x)$ holds. This is true because $1-\frac{2x}{\pi}≥1≥\cos(x)$ for all $x\in\left(-\infty;0\right]$

So the inequality is indeed true.