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

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.)

Solution 4:

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 $e^x \leq x+e^{x^2}$.