Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does anybody have a simple proof this inequality $$e^x\le x+e^{x^2}.$$


share|cite|improve this question
up vote 15 down vote accepted

For $x \geq 0$, $$ e^{x^2 - x} + x e^{-x} \geq 1 + (x^2 - x) + x (1-x) = 1 \> . $$

For $x < 0$, let $y = - x > 0$, whence, $$ e^{-x^2}(e^x - x) = e^{-y^2} (e^{-y} + y) \leq e^{-y^2} (1 - y + y^2/2 + y) \leq e^{-y^2}(1+y^2) \leq e^{-y^2} e^{y^2} = 1 \> . $$

Notice that we've only used the basic facts that $e^x \geq 1 + x$ for all $x \in \mathbb{R}$ and that $e^{-x} \leq 1 - x + x^2/2$ for $x \geq 0$, both of which are trivial to derive by simple differentiation, similar to Didier's approach.

share|cite|improve this answer
I like your approach the most! – Qiang Li Mar 5 '11 at 21:41

Consider $f(x) = x+\mathrm{e}^{x^2}-\mathrm{e}^x$. Then $f'(x) = 1+ 2x\mathrm{e}^{x^2}-\mathrm{e}^x$ and $$f''(x) = 2(1+2x^2)\mathrm{e}^{x^2}-\mathrm{e}^x\ge \mathrm{e}^xg(x), $$ with $g(x)=2\mathrm{e}^{x^2-x}-1$. Since $x^2-x\ge-\frac14$, $g(x)\ge2\mathrm{e}^{-1/4}-1>0$ for every $x$, hence $f''$ is positive everywhere and $f'$ is increasing. Since $f'(0)=0$, $f'(x)<0$ if $x<0$ and $f'(x)>0$ if $x>0$. Thus, the function $x\mapsto f(x)$ is decreasing on $x\le0$ and increasing on $x\ge0$. Since $f(0)=0$, we are done.

share|cite|improve this answer

Hint: Consider the power series of $e^x$ which converges on all of $\mathbb{R}$. Then $$e^x-x=1+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$$ and $$e^{x^2}=1+x^2+\frac{x^4}{2}+\frac{x^6}{3!}+\cdots .$$ For small $x$ the $x^2$ dominates, and for $|x|\geq 1$ the second power series will dominate the first. (The coefficient of $x^{2n}$ is large enough to cover both coefficients of $x^{2n-1}$ and $x^{2n}$ in the first expansion.)

share|cite|improve this answer
but is this rigorous, comparing two infinite sums? – Qiang Li Mar 5 '11 at 19:26
Yes, it is perfectly fine since these series converge nicely everywhere. – Eric Naslund Mar 5 '11 at 19:30
If I may complete your post, the coefficient of $x^{2n}$ in the expansion of $\mathrm{e}^{x^2}$ is indeed larger than the sum of the coefficients of $x^{2n−1}$ and $x^{2n}$ in the expansion of $\mathrm{e}^x-x$, hence the conclusion holds for $|x|\ge1$. But this argument cannot work for $|x|<1$. Fortunately, the same coefficient of $x^{2n}$ in the expansion of $\mathrm{e}^{x^2}$ is also larger than the sum of the coefficients of $x^{2n+1}$ and $x^{2n}$ in the expansion of $\mathrm{e}^x-x$, hence the conclusion holds for $|x|\le1$ as well. – Did Mar 6 '11 at 9:05
@Didier: Thanks, that is a nice way of doing it! – Eric Naslund Mar 6 '11 at 16:32

Consider $f(x) = x+e^{x^2}-e^x$. Then $f'(x) = 1+ 2xe^{x^2}-e^x$. Find the critical points. So the minimum of $f(x)$ is $0$ which implies that $e^x \leq x+e^{x^2}$.

share|cite|improve this answer
how did you find the critical points then? – Qiang Li Mar 5 '11 at 19:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.