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I'm trying to prove the following inequality: $$e^{x^2}+x\geq e^x$$ I can prove it for $x<0$ and for $x>1$, but I'm having trouble with $x\in[0,1]$.
I've tried to integrate both sides $$\int_0^1(e^{x^2}+x) dx\geq \int_0^1e^x dx\iff\int_0^1e^{x^2} dx\geq e-\frac{3}{2}$$ but I can't integrate $\int_0^1e^{x^2}dx$.
I feel like I'm missing some other simpler ways. Could someone help?

Francesco Virgili
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    Note: $e^{x^2}$ doesn't have an elementary integral, so there's no hope of writing down a formula for that antiderivative. Incidentally, that particular antiderivative (up to a scalar) is important enough that it has a name: the imaginary error function. – Sammy Black Jul 26 '22 at 22:19
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    Try substituting Taylor series; use the fact that $x^p > x^q$ for $p < q$ for $x \in (0,1).$ – William M. Jul 26 '22 at 22:19
  • Do you know how to use the Mean Value Theorem to establish inequalities like this? – Sammy Black Jul 26 '22 at 22:26
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    @WangYeFei In that link, cardinal's proof for $x \ge 0$ is very nice. For $x < 0$, alternative proof: Using the well-known $\mathrm{e}^u \ge 1 + u$, we have $\mathrm{e}^{x^2} \ge 1 + x^2$ and $\mathrm{e}^x = \frac{1}{\mathrm{e}^{-x}} \le \frac{1}{1 - x}$. Thus, we have $\mathrm{LHS} - \mathrm{RHS} \ge 1 + x^2 + x - \frac{1}{1 - x} = \frac{-x^3}{1 - x} \ge 0$. – River Li Jul 26 '22 at 22:57
  • A proof using Maclaurin series: For $x\in [0, 1]$, we have $\mathrm{e}^x = 1 + x + \sum_{k=2}^\infty \frac{x^k}{k!} \le 1 + x + \sum_{k=2}^\infty \frac{x^2}{k!} = 1 + x + x^2\left(\sum_{k=0}^n \frac{1}{k!} - 2\right) = 1 + x + x^2(\mathrm{e} - 2)$. Also, $\mathrm{e}^{x^2} \ge 1 + x^2$. – River Li Jul 26 '22 at 23:06

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Compute the Maclaurin series.

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots $$

$$ e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \ldots $$

So,

$$ e^x - e^{x^2} = x - \frac{x^2}{2!} + \frac{x^3}{3!} - \left( \frac{1}{2!} - \frac{1}{4!} \right) x^4 + \ldots $$

which is an alternating series of decreasing terms if $x \in [0,1],$ and since the first term is $x$ and the next term is non-positive, it follows that $ e^x - e^{x^2} \leq x\ \forall x \in [0,1].$

Adam Rubinson
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