How can I solve the inequality $e^{2x^2 + x} \geq \frac{1}{\sqrt{1-2x}}$

Question

I want to show that the following inequality holds for all $|x|<1/4$

$$e^{2x^2 + x} \geq \frac{1}{\sqrt{1-2x}}$$

How can I do that? I tried Wolfram alpha and the above inequality is correct for $x \leq 0.34$.

Here is my attempt:

We know that $e^y > 1+y$. We have

$$e^{4x^2 +2x } > 4x^2 + 2x + 1$$

If we solve for $4x^2 + 2x + 1 \geq \frac{1}{1-2x}$, we have $(1+2x)^2(1-2x) \geq 1$ which is clearly satisfied for $0<x<1/4$. However, this inequality is not satisfied for $x<0$.

Answer 1

Hint: Consider $(1-2x)e^{4x^{2}+2x}$. It is easy to see that in $(-\frac 1 4, \frac 1 4)$, the derivative of this function is $0$ only at $x=0$. Hence it is enough to show that $(1-2x)e^{4x^{2}+2x} \geq 1$ at $0$ and the end points $\pm \frac 1 4$. This is quite easy.

In fact the function is decreasing in $(-\frac 1 4,0)$ and increasing in $(0,\frac 1 4)$. Its value at $x=0$ is $1$. Hence the function is $\geq 1$ at all points.

Answer 2

We need to prove that $f(x)\geq0,$ where $-\frac{1}{4}<x<\frac{1}{4}$ and $$f(x)=2x^2+x+\frac{1}{2}\ln(1-2x).$$ Indeed, $$f'(x)=4x+1-\frac{1}{1-2x}=\frac{2x(1-4x)}{1-2x},$$ which gives $$x_{min}=0,$$ $$f(x)\geq f(0)=0$$ and we are done!