Is $\cos(x) \geq 1 - \frac{x^2}{2}$ for all $x$ in $R$?

Consider the function $$ f(x)=\cos x - 1 + \frac{x^2}{2} $$ We have $f(0)=0$ and $$ f'(x)=x-\sin x $$ with $f'(0)=0$. Also $$ f''(x)=1-\cos x $$ which shows that $f'(x)$ is a strictly increasing function, because its derivative is positive except on a set of isolated points (that has no limit point). Therefore $f'(x)>0$ for $x>0$ and $f'(x)<0$ for $x<0$. Hence $0$ is an absolute minimum for $f$. Since $f(0)=0$, we have $f(x)>0$ for every $x\ne0$. This means that, for every $x$, $$ \cos x\ge 1-\frac{x^2}{2} $$ equality holding only for $x=0$.

Another method is using the well known inequality $\sin { x\le x } $ and integrating both sides $$\int _{ 0 }^{ x }{ \sin { tdt\le \int _{ 0 }^{ x }{ tdt } } } $$ $${ \cos { t } | }_{ 0 }^{ x }\le { \frac { { t }^{ 2 } }{ 2 } | }_{ 0 }^{ x }$$ $$-\cos { x } +1\le \frac { { x }^{ 2 } }{ 2 } \\$$

$$ \cos { x } \ge 1-\frac { { x }^{ 2 } }{ 2 } $$