Proof of $ e^x-x^2 \gt 1 $ when $ x \gt 0$ and $x$ is a real number .

Question

I want to Prove $ e^x-x^2 \gt 1 $ when $ x \gt 0$ and $x$ is a real number . For this purpose , my trying is as the following :

$ e^x-x^2 = \{1+x+\dfrac{x^2}{2!} + +\dfrac{x^3}{3!}++\dfrac{x^4}{4!}+ ....... \}-x^2$ $= 1+\{x-\dfrac{x^2}{2!} + +\dfrac{x^3}{3!}++\dfrac{x^4}{4!}+ ....... \}$

So $ e^x-x^2 \gt 1 $ is true if $\{x-\dfrac{x^2}{2!} + +\dfrac{x^3}{3!}++\dfrac{x^4}{4!}+ ....... \} \ge $ Please help me finding the convergence of this term .

Answer 1

The second derivative of $e^x-x^2-1$ gives you the inflection point at $x = ln(2)$. The problem is equivalent to $e^{ln(2)}-len(2)^2>1$ or $1>ln(2)$, wich is true