# Mean Value Theorem Inequality Contradiction(?)

I am trying to show:

$e^x > 1+x+\frac{x^2}{2}$ for $x>0$ using Mean Value Theorem (MVT).

My method is as follow: consider the function $f(x)=e^x - (1+x)$. We know $f'(x) = e^x -1 > 0$ for $x>0$. Thus, by MVT, exists a $c \in ]0,x[$ such that, $$f'(c)=\frac{f(x)-f(0)}{x-0}=\frac{e^x-(1+x)}{x}$$ Since we know $f'(c) > 0$, the statement is equivalent to, $$\frac{e^x -(1+x)}{x} > 0 \Rightarrow e^x - 1> x \Leftrightarrow f'(x)>x$$ for $x>0$. Now we can update the original inequality of $f'(c)$, giving $$f'(c)=\frac{e^x-(1+x)}{x} > x \Leftrightarrow e^x > 1+x+x^2$$ for $x>0$. However, this is not the same as the intended inequality, and is certainly not true, letting $x=1$ yields $e > 3$. What happened here? May someone explain?

Thank you so much!

• You can only get $f'(x)>x$, but not $f'(c)>x,\forall c\in]0,x[$. – Eclipse Sun Apr 8 '15 at 9:17

You did wrong in your last step. You proved $f'(x) > x$, but write in the next line $f'(c) > x$. You have (as $x$ was arbitrary), $f'(c) > c$, giving $$e^x > 1 + x + cx$$ but this does not help.

Different approach: consider $h(x) = e^x -1 -x -\frac{x^2}{2}$. Clearly $h(0)=0$. Now show that $h'>0 \forall x>0$ thus proving the inequality.

It is universally known that $e^x> 1+x$ for $x> 0$. Thus $$e^x=\left(e^{x/n}\right)^n> \left(1+\frac{x}{n}\right)^n> 1+n·\frac{x}{n}+\frac{n(n-1)}2·\frac{x^2}{n^2}=1+x+\left(1-\frac1n\right)·\frac{x^2}2.$$ Since this is true for all $n>2$, the requested inequality follows.