Proof of the inequality $e^x\le e^{x^2} + x$

Question

The question is to prove the inequality $e^x\le e^{x^2} + x$. I tried the Taylor expansion like ${e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ...$ and $x + {e^{{x^2}}} = 1 + x + \frac{{{x^4}}}{{2!}} + \frac{{{x^6}}}{{3!}} + ...$ but cannot see anything useful out of this. Anyone can provide some help? Thank you.

Answer 1

Note that for any $x$, we have $$ (e^{x^2} + x) - e^x = \\ \frac{1}{2!}x^2 - \frac{1}{3!}x^3 + \left(\frac{1}{2!} - \frac{1}{4!}\right)x^4 - \frac 1{5!}x^5 + \left( \frac{1}{3!} - \frac{1}{6!} \right)x^6 - \cdots \geq \\ \frac{1}{2!}x^2 - \frac{1}{3!}x^3 + \frac {1}{4!}x^4 - \frac 1{5!}x^5 + \frac{1}{6!}x^6 - \cdots =\\ e^{-x} + x - 1 $$ Consider the function $f(x) = e^{-x} + x - 1$. We note that $$ f'(x) = 1 - e^{-x} $$ so that $f$ achieves its minimum at $x = 0$, where we find $f(0) = 0$. So, $f(x) \geq 0$.

Thus, we have $$ (e^{x^2} + x) - e^x \geq f(x) \geq 0 $$ The conclusion follows.

Answer 2

One other way is to consider $f(x)=e^{x^2}+x-e^x$, we have $f'(x)=2xe^{x^2}+1-e^x$ and $f''(x)=(4 x^2+2) e^{x^2}-e^x$, So $$f''(x)=e^x\left(e^{x^2-x}(4x^2+2)-1\right)\ge e^x\left(e^{-1/4}\times 2-1\right)>0$$ Since $x^2-x\ge-1/4$. So, $f$ is convex, and because $f'(0)=0$ we conclude that $f$ attains its minimum at $x=0$ which is $0$. The desired inequality is proved.

Answer 3

If $x \geq 1$, then $x^2 \geq x$, we then have $e^x \leq e^{x^2}$, which in turn means that $e^x \leq e^{x^2}+x$.

For $0 \leq x \leq 1$, then $x^2 \leq x$, we then have $e^x$, we then have \begin{align} e^x & = \sum_{k=0}^{\infty} \dfrac{x^k}{k!} = 1 + x + \sum_{k=2}^{\infty} \dfrac{x^k}{k!} \leq 1 + x + \sum_{k=2}^{\infty} \dfrac{x^2}{2^{k-1}} \,\,\,\, (\because x \in [0,1] \text{ and }k! \geq 2^{k-1})\\ & = 1+x+x^2 \leq x + e^{x^2} \end{align}

If $x \leq 0$, then $e^{x^2}+x \geq 1+x+\dfrac{x^2}2 = \dfrac{1+(1+x)^2}2 \geq e^x$.

Hence, $e^x \leq e^{x^2}+x$ for all $x \in \mathbb{R}$.