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 $$\int _{ 0 }^{ x }{ \sin { tdt\le \int _{ 0 }^{ x }{ tdt } } } $$ $${ \cos { t } | }_{ 0 }^{ x }\le { \frac { { t }^{ 2 } }{ 2 } | }_{ 0 }^{ x }$$ $$-\cos { x } +1\le \frac { { x }^{ 2 } }{ 2 } \\$$

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

Consider the definition of $\cos x$ as the sum of a power series: $$\cos x=\sum_{k\ge0}(-1)^k\frac{x^{2k}}{(2k)!}.$$ As $\cos x$ is an even function, and the series has terms of even degree only, we may suppose $x\ge 0$.

Furthermore, we may suppose $x\le 2$ because $1-\frac{x^2}2<-1$ for $x>2$. For each $x\in[0,2]$, it's an alternating series with terms decreasing in magnitude. Partial sums are approximations of $\cos x$ with an error $$\cos x-\biggl(\sum_{k=0}^n(-1)^k\frac{x^{2k}}{(2k)!}\biggr)= (-1)^{n+1}\frac{x^{2(n+1)}}{(2(n+1))!}+\dotsm$$ which has the sign of the first omitted term (and its absolute value is not more than the absolute value of this term).

Hence $\;1-\dfrac{x^2}{2}<\cos x$.