TITLE: Using mean value theorem to show that $\cos (x)>1-x^2/2$
QUESTION [6 upvotes]: I have a question, by applying the mean value theorem  to $f(x)=\frac{x^2}{2}+\cos (x)$, on the interval $[0,x]$, show that $\cos (x)>1-\frac{x^2}{2}$.
We know that $\frac{\text{df}(x)}{\text{dx}}=x-\sin (x)$, for $x>0$. By the MVT, if $x>0$, then $f(x)-f(0)=(x+0) f'(c)$ for some $c>0$.
This is where I get confused:
so, $f(x)>f(0)=1$, but why? Is it my lack of inequality that is showing, or what am I missing? Is $f'(x)\cdot x=1$ or what is going on?

REPLY [8 votes]: $ f(x)=x^2/2+\cos(x)$
Note that $f(0)=0^2/2+1=1$
From your equation:
$$f(x)-f(0)=(x)f'(c)=x(c-\sin c)$$

Let $g(x)=x-\sin x$
  Again you can show that $g'(x)=1-\cos x$ which is always greater than
  $0$ due to bounded nature of $\cos x$.As $g(0)=0$ and it is an
  increasing function $\{g'(x)>0\;\forall x>0$}, thus $g(x)>0 \;\forall x>0$.

So $f(x)-f(0)>0\;\forall x>0$ as $x>0$ and $c-\sin c >0\;\forall c>0${as $0<c<x$}.
So $f(x)>f(0)=1$