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How it can be shown that:

$$e^{x}<1+x+x^{2}$$ For all $x<0.5$ I tried to use mean value theorem, but I have some problem, any idea or hint if highly appreciated. Clearly $e^x$ is continuous over $\left[x,x+1\right]$ and differentiable over $\left(x,x+1\right)$, hence there exist $c∈\left(x,x+1\right)$, such that: $$\frac{f\left(x+1\right)-f\left(x\right)}{x+1-x}=f^{'}\left(c\right)$$ Hence; $$e^{\left(x+1\right)}-e^{\left(x\right)}=e^c$$ or $$\ln\left(e^{\left(x+1\right)}-e^{\left(x\right)}\right)=c$$ but this is not helpful.

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    $\begingroup$ What are your assumptions about $x$? The strong inequality is wrong when $x=0$. $\endgroup$ – Christian Blatter Dec 5 '19 at 14:43
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    $\begingroup$ $e^x = \sum_{k=2}^{\infty}\frac{x^k}{k!}+x+1< \sum_{k=2}^{\infty}\frac{x^2}{k!}+x+1 = x^2(e^1 - 2)+x+1)<x^2+x+1$ $\endgroup$ – fGDu94 Dec 5 '19 at 14:43
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This is not a complete answer, but I thought it provides a simple method to find some values of x.

That expression is not true for all $x<0.5$ . Try $0$, and you obtain $1>1$ . For $x<-0.5$ the inequality holds, because we can write $e^x -1 < x(x+1)$ and if we differentiate we obtain $e^x$ and $2x+1$. Then, for all $x<-1/2$, the derivative of $e^x -1 $ is positive while the one for $x(x+1)$ is negative. Since $e^{(-1/2)} - 1 < 1+1/2+(1/2)^2$ then the inequality holds for $x< - 0.5$.

Actually, it is true for $ x \in ]-∞;0[ \; \cup \; ]0;1.79328[$ , but I'm not sure how to prove it analytically, neither of where the $1.79328$ comes from (aproximated value). I hope my method is close enough to what you wanted.

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  • $\begingroup$ the inequality is equivalent to $e^x-x^2-x-1<0$. So the $1.79328$ is one of the roots of $e^x-x^2-x-1$ $\endgroup$ – Zacharias Zarowski Dec 5 '19 at 16:42
  • $\begingroup$ Yeah, I knew, I just don't know a algebraic expression for it. But one could express it that way, yes. $\endgroup$ – RicardoMM Dec 5 '19 at 17:55
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    $\begingroup$ Ok I'm also not sure how to find a closed form :) but in this case its enough since its greater than $0.5$ $\endgroup$ – Zacharias Zarowski Dec 5 '19 at 18:31
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Let us consider the function $$ g(x) = e^x - 1 - 2x $$ We have that $g(0)=0$. Moreover, since $g(1)=e-3<0$ while $g(2)=e^2-5>0$, by the theorem of zeros, there is a value $1<a<2$ such that $g(a)=0$. Since the function $f(x)=e^x$ is strictly convex, it must be $e^x<1+2x$ or each $0<x<a$. Therefore $$ \int\limits_0^x {e^t dt} < \int\limits_0^x {\left( {1 + 2t} \right)dt} $$ for each $0<x<a$. It means that $$ e^x - 1 < x + x^2 $$ for such values of $x$. This prove the inequality for $0<x<0.5$. Now, let be $x<0$. Then, again by convexity, it is $e^t>1+t$ for each $t\in \mathbb R$, $t\neq 0$. Therefore $$ \int\limits_x^0 {e^t dt} > \int\limits_x^0 {\left( {1 + t} \right)dt} $$ so that $$ 1 - e^x > - x - \frac{{x^2 }} {2} $$ thus $$ 1 + x + \frac{{x^2 }} {2} > e^x $$ and, a fortiori, the inequality follows.

You can also prove that $ e^t <1+2t $ for each $ 0< t< 1/2$ also by means of mean value theorem. Whith $0 < t <1/2$ you have that $$ \frac{{e^t - e^0 }} {{t - 0}} = e^c < e^{\frac{1} {2}} < 2 $$ thus $e^t< 1+2t$ if $0 < t < 1/2$.

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Two partial integrations give the following version of Taylor's theorem: $$\eqalign{e^x&=1+\int_0^x 1\cdot e^t\>dt=1+(t-x)e^t\biggr|_{t=0}^{t=x}-\int_0^x(t-x)e^t\>dt\cr &= 1+x-{(t-x)^2\over2}e^t\biggr|_{t=0}^{t=x}+\int_0^x{(t-x)^2\over2}e^t\>dt\cr &=1+x+{x^2\over2}+\int_0^x{(t-x)^2\over2}e^t\>dt\ .\cr}$$ We therefore have to show that $$\int_0^x{(t-x)^2\over2}e^t\>dt\leq{x^2\over2}\qquad(-\infty<x\leq1)\ .$$ When $x\leq0$ this is obviously true. When $0<x\leq1$ we have $$\int_0^x{(t-x)^2\over2}e^t\>dt\leq{e\over2}\int_0^x\tau^2\>d\tau={e\over2}{x^2\over3}<{x^2\over2}\ .$$

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