Let $X$ be a random variable and $X\sim N(0,1)$. Prove that: $\mathbb{P}(-R\leq X\leq R)\geq \sqrt{1-e^{\frac{-R^2}{2}}}$ for $R>0$

Question

Let $X$ be a random variable and $X\sim N(0,1)$. Prove that: $\mathbb{P}(-R\leq X\leq R)\geq \sqrt{1-e^{\frac{-R^2}{2}}}$ for $R>0$. I wrote the left side of the inequation as either $2F(R)-1$ or $1-2F(-R)$ where $F$ is the quantile function of normal distribution, but from there I can't find a lower estimation that would lead me to the right side of the inequality. I've also tried to integrate it, but I couldn't do it alone and integral calculator returned the Gauss Error Function, which I'm not familiar with, and so I'm not able to apply it here.

Answer 1

Notice, that if $Y$ is independent of $X$, with $Y\sim X \sim N(0,1)$, then $$\mathbb{P}(-R\leq X \leq R)^2 = \mathbb{P}(-R\leq X \leq R,-R\leq Y \leq R)\geq \mathbb{P}(X^2+Y^2 \leq R^2).$$ The right hand side can be integrated using polar coordinates \begin{align*} \mathbb{P}(X^2+Y^2 \leq R^2) &= \frac{1}{2\pi}\int\int_{x^2+y^2\leq R^2}e^{-(x^2+y^2)/2} \: dxdy \\ &= \frac{1}{2\pi} \int_0^{2\pi}\int_0^R re^{-r^2/2} \: drd\theta \\ &= \int_0^R re^{-r^2/2} \: dr \\ &= 1- e^{-R^2/2}, \end{align*} which immediately gives the desired result. Notice how the above computation is quite similar to how one usually evaluates the Gaussian integral.