How to show that $e^x\le x + e^{x^2}$ for any real $x $

Question

Show that $e^x\le x + e^{x^2}$ for any real $x$.

I have trouble proving this inequality for $x \in (0, 1)$. Any hint?

Answer 1

Assume that $f(x)=e^x-x-e^{x^2}$ then find $f'(x)$ and show that this is less than $0$ for $x\in(0,1)$.

Answer 2

HINT: define $$f(x)$$ as $$f(x)=x+e^{x^2}-e^x$$ and discuss this function with the help of calculus.

Answer 3

We prove that $e^{x^2}+x\geq e^x$ for all $x\in\mathbb{R}$, with equality iff $x=0$.

Let $f(x)=e^{x^2}+x-e^x.$ First calculate $f'(x)=1+2xe^{x^2}-e^x,$ $f''(x)=(4x^2+2)\,e^{x^2}-e^x.$ Note that $f''(x)\geq0$ iff $\ln(4x^2+2)+x^2-x\geq0,$ but $$\ln(4x^2+2)+x^2-x\geq\ln2-\frac{1}{4}>0,$$ so $f'(x)$ is monotonically increasing. Hence $f'(x)>f'(0)=0$ for $x>0$, and $f'(x)<0$ for $x<0$. Thus $f(x)$ is monotonically decreasing on $(-\infty,0)$ and increasing on $(0,\infty)$, attaining its local (and global) minimum at $x=0$, where $f(0)=0$.

Answer 4

Let $f(x)=x+e^{x^2}-e^x$, $f(0)=0$

$f'(x)=1+2xe^{x^2}-e^x$, $f'(0)=0$

$f''(x)=2e^{x^2}+4x^2e^{x^2}-e^x=e^x(2e^{x^2-x}-1)+4x^2e^{x^2}$

$e^{x^2-x}=e^{(x-\frac{1}{2})^2-\frac{1}{4}}\ge e^{-\frac{1}{4}}>\frac{1}{2}$, so $f''(x)>0$ for $x\in(0,1)$, done.

Another try:

By Taylor expansion about $x=0$, $e^x=1+x+\frac{x^2}{2}+R_3(x)$

$R_3(x)=\frac{e^c}{3!}x^3\le \frac{e}{6}x^3$

$x+e^{x^2}=x+(1+x^2+\cdots)>1+x+x^2$

$\frac{1}{2}>\frac{e}{6}\ge \frac{e}{6}x \Rightarrow \frac{1}{2}x^2>\frac{e}{6}x^3 \Rightarrow 1+x+x^2>1+x+\frac{x^2}{2}+R_3(x)\Rightarrow x+e^{x^2}>e^x$