Prove that If $0
If $0<x<$ln $2$ then $x^2\geq e^x-x-1$
I got this problem while reading a proof. Tried to prove it but failed. $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$. So $e^x\geq 1+x$ for all $x$ but how can I involve $x^2$ here? Can anyone please give me a hint.
 A: Consider $f(x)=x^2-e^x+x+1$. Note that $f(0)=0$ and $f'(x)=2x-e^x+1$ also satisfies $f'(0)=0$. Moreover, $f''(x)=2-e^x\geq 0$ for $x\in[0,\log(2)]$. All this implies $f'(x)\geq 0$ for $x\in[0,\log(2)]$. The claim follows.
A: Just another way, by integration
$$x<\ln 2 \implies e^x<2\implies \int_0^xe^t \,dt < \int_0^x  2 \, dt \implies e^x-1 < 2 x \implies \int_0^x(e^t-1)\,dt < \int_0^12t\,dt \implies e^x-x-1< x^2$$
A: We have to prove that over $I=(0,\log 2)$ we have $e^x\leq 1+x+x^2$.
Since $e^x$ is a convex function,
$$ \forall z\in I,\qquad e^z \leq 1+\frac{z}{\log 2} $$
and by integrating both sides over $[0,x]$,

$$ \forall x\in I, \qquad e^{x} \leq 1+x+\frac{x^2}{2\log 2} $$

follows. The last inequality is stronger than the inequality we had to prove, since $2\log 2>1$, or $e<4$.
A: As there seems to be some uncertainty regarding the lower bound, I'll write out the proof via the standard formula for the remainder.  Suppose x $\in$ (-$\infty$,$\ln 2$). Then Taylor's Theorem with Remainder tells us that there exists some real number $c \in(-\infty,x]$ such that $$e^x \;= 1 + x + \frac{e^c}{2}x^2$$  Clearly $c < \ln 2$ so $e^c < 2$ hence $e^x < 1+x + x^2$.  The claim follows.  Note that at no point did any lower bound on x appear.
A: We can prove the stronger

$$0\le x\le1 \implies x^2\geq e^x-x-1$$

indeed let consider $f(x)=e^x-1-x-x^2\le0,\, x\in[0,1]$ then


*

*$f(0)=0$

*$g(x)=f'(x)=e^x-1-2x  \le 0$
indeed 


*

*$g(0)=0$

*$g(1)=-2$

*$g'(x)=e^x-2=0 \implies x=\ln 2$ which is a minimum


therefore the inequality holds.
