x=e^t

therefore

g(t)=( 1/2)(e^t+e^(-t))

h(t)=( 1/2)(e^t-e^(-t))

f(t)=g(t)+h(t)=e^t

f(x)=e^x

Let f (x) = e x-1-x-x 2/2.

F(0)= 1- 1-0-0=0

F'(x)=e^x- 1-x

F'(x)>0 to x>0 holds.

Because F'(0)= 1- 1-0=0.

The next requirement is that f'' (x) = e x- 1 > 0 to x>0 holds.

Obviously, because e x is increasing, when x=0, the function value is 1, x >；; E x of 0 must be greater than 1

So F''(x) is constant and increasing when x >: 0, so F'(x) >; 0

So 0 in F(x)>x >; 0 Shangheng is established.

So f (x) >1+x+x 2/2 in x > 0 Shangheng is established.