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.

a) You'r right, you can do this with the discriminant and it is very natural. But you can also use the well-known inequality: $2ab\leq a^2+b^2$ which follows from the expansion of $(a-b)^2\geq 0$. So you get
$$
2x=2\cdot x\cdot 1\leq x^2+1^2=x^2+1<x^2+2\qquad \forall x\in\mathbb{R}.
$$
Then divide by $2$.
b) By definition, $\cosh x$ is the even part of $e^x$. Since $e^x=\sum_{n\geq 0}\frac{x^n}{n!}$ for every $x$, you are left with
$$
\cosh x=\sum_{n\geq 0}\frac{x^{2n}}{(2n)!}=1+\frac{x^2}{2}+\frac{x^4}{24}+\ldots\qquad\forall x\in\mathbb{R}.
$$
Now, using  inequality a) and the fact that every term of the series is nonnegative:
$$
\cosh x=1+\frac{x^2}{2}+\frac{x^4}{24}+\ldots\geq 1+\frac{x^2}{2}> x\qquad\forall x\in\mathbb{R}.
$$
c) The distance from the point $(x_0,y_0)$ to the line $x-y=0$ is 
$$
d=\frac{|x_0-y_0|}{\sqrt{1^2+(-1)^2}}=\frac{1}{\sqrt{2}}|x_0-y_0|.
$$
So for a point $(x,\cosh x)$ on the graph of the hyperbolic cosine, we get
$$
d(x)=\frac{1}{\sqrt{2}}|x-\cosh x|=\frac{1}{\sqrt{2}}(\cosh x-x)
$$
as $\cosh x > x$ for every $x$. Now study the variations of $d(x)$ to find its minimum. It will occur at a critical point $d'(x_0)=0$. And actually, there is only one such critical point, solution of $\sinh x=1$. So your minimum occurs at
$$
x_0=\mbox{arsinh} 1=\log (1+\sqrt{2})\qquad y_0=\cosh x_0=\sqrt{2}.
$$
Click on this link for a drawing of the situation.
