Hyperbolic cosine I have an A level exam question I'm not too sure how to approach: 
a) Show $1+\frac{1}{2}x^2>x, \forall x \in \mathbb{R}$
b) Deduce $ \cosh x > x$
c) Find the point P such that it lies on $y=\cosh x$ and its perpendicular distance from the line $y=x$ is a minimum.
I understand how to show the first statement, by finding the discriminant of $1+\frac{1}{2}x^2-x>0$, but trying to apply this to part b doesn't seem to work:
$$\cosh x > x \Rightarrow \frac{1}{2}(e^x+e^{-x}) > x \Rightarrow e^x+e^{-x}>2x \Rightarrow e^{2x}-2xe^x+1>0$$
Applying $\Delta>0$ to this because I know the two functions do not intersect:
$$\Delta>0 \Rightarrow 4x^2-4>0 \Rightarrow x^2>1 \Rightarrow |x|>1$$
This tells me that in fact, the two do meet, but only when $|x|>1$, what have I done wrong here?
This last bit I don't know how to approach, I was thinking maybe vectors in $\mathbb{R}^2$ were involved to find perpendicular distances?
Thanks in advance.

For $(a)$ just consider the function
$$ y = 1+\frac{x^2}{2}-x \implies y-\frac{1}{2}= \frac{1}{2}(x-1)^2 $$ 
which is a parabola with vertex at $(1,\frac{1}{2})$ and implies that 
$$ y > 0 \implies 1+\frac{x^2}{2}-x>0 \implies 1+\frac{x^2}{2}>x, \forall x \in \mathbb{R} .$$
