Prove that $f(x) := e^{x^2/2}-x \ge 0$ for all $x$

Question

Define $f:\mathbb R_+ \to \mathbb R$ by $f(x) := e^{x^2/2}-x$. I would like to prove that $f(x) \ge 0$ for all $x \ge 0$.

1. Could you have a check on my attempt?
2. Are there other (more direct) approaches?

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We have $f'(x) = xe^{x^2/2}-1$. Then $f'(0)=-1$ and $f'$ is strictly increasing. So there is a unique $x_0 \ge 0$ such that $f'(x_0)=0$. Clearly, $f$ attains its minimum at $x_0$. So it suffices to prove $f(x_0) \ge 0$. We have $f'(x_0)=0 \iff e^{x_0^2/2} = 1/x_0$. Then $$ f(x_0) = \frac{1}{x_0} - x_0. $$

Because $f'(1) \approx 0.65>0$, we get $x_0 < 1$. Then $f(x_0) >0$. This completes the proof.

Answer 1

Your approach is good, but it does not explain how you got $f'(1)\approx0.65$. You only want to show that $x_0<1$, for which it suffices to show that $f'(1)>0$ and this is quite clear from the fact that $$f'(1)=1\cdot e^{1^2/2}-1=\sqrt{e}-1.$$ Apart from this your proof seems very complete. You could mention explicitly that $f'$ is strictly increasing because the domain of $f'$ is $\Bbb{R}_{+}$.

Answer 2

There is a well known inequality that $e^x\geq 1+x$ for all $x$ for example by convexity of the exponential function.

In particular $e^{x^2/2}\geq 1+\frac{x^2}{2}$. But $1+\frac{x^2}{2}\geq x$ since $2+x^2-2x = (x-1)^2+1\geq 0$ for all $x$.