Elementary proof of an inequality with $e^x$ when $|x|<1$. Assume $|x| <1 $ and we already know $0 \leq e^x - 1 - x$. Note that last inequality was god given, we know it is true, but we do not know how it was proved. Can we deduce from here that $e^x - 1 - x \leq x^2$ without using derivatives etc., just only with elementary algebraic manipulations and ideas of convexity? 
 A: If $f$ is convex and differentiable, we have $f(x)-f(y) \ge f'(y) (x-y)$ for all $x,y$.
The function $\exp$ is convex and differentiable, so we have $e^x-1 \ge x$ for all $x$.
Let $\phi(x) =e^x-1-x$, then $\phi(x) \ge 0$, $\phi(0)=\phi'(0) = 0$ and $\phi''(x) = e^x$.
The mean value theorem gives $\phi(x) = \phi(0) + \phi'(0) x + {1 \over 2} \phi''(c) x^2$ for some $c \in (0,x)$. Since $e^x \le 1$ for $x \le 0$, we have
$\phi(x) \le {1 \over 2} x^2$ for all $x \le 0$.
My estimate for $x \in (0,1]$ is messier:
Assuming $x>0$, we have $\phi(x) \le x^2$ iff ${ x^2 \over 2!} + { x^3 \over 3!} + \cdots \le x^2$ iff
${ x^3 \over 3!} + \cdots \le { x^2 \over 2!} $ iff
${ x \over 3!} + { x^2 \over 4!} +\cdots \le { 1 \over 2!} $.
We have the estimate
${ x \over 3!} + { x^2 \over 4!} +\cdots = { x \over 3!} (1 + { x \over 4} + {x^2 \over 4 \cdot 5 } + \cdots ) \le { x \over 3!} (1 + { x \over 4} + {x^2 \over 4^2 } + \cdots ) = { x \over 3!}{ 1 \over 1-{x \over 4}} \le {2 \over 9} x$, where we have used the fact that ${ 1 \over 1-{x \over 4}} \le {4 \over 3}$ when $x \in (0,1]$. Since ${2 \over 9} x \le {1 \over 2}$ for all
$x \in (0,1]$ we obtain $\phi(x) \le x^2$ for all $|x|<1$.
