Answer
$coshx\geq 1+\frac{x^{2}}{2}$
Work Step by Step
As we know, $$coshx=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}$$
This series converges absolutely for $x$ ,
Since, $\lim\limits_{n \to \infty}|\frac{\frac{x^{2n+2}}{(2n+2)!}}{\frac{x^{2n}}{(2n)!}}|=\lim\limits_{n \to \infty}|\frac{x^{2}}{(2n+1)(2n+2)}|=0\lt1 $
and thus, by ratio test the series absolutely converges for $x$
Hence, $coshx=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}=1+\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+...\geq 1+\frac{x^{2}}{2!}$
Thus, $coshx\geq 1+\frac{x^{2}}{2}$
