Prove $\frac{-4 \sqrt x + 2 e^x \sqrt x + \sqrt \pi \operatorname{erfi}\sqrt x}{2 \sqrt x}\leq \frac{e^{3x}-3x-1}{3x}$.

Question

From Wikipedia, the imaginary error function, denoted erfi, is defined as $$\operatorname{erfi}(x) = \frac{2}{\sqrt\pi} \int_0^xe^{t^2}\,\mathrm dt.$$ Prove that $$\frac{-4 \sqrt x + 2 e^x \sqrt x + \sqrt \pi \operatorname{erfi}\sqrt x}{2 \sqrt x}\leq \frac{e^{3x}-3x-1}{3x}$$ for all $x>0$.

Answer 1

An obvious first step is to get rid of all the square roots by letting $\sqrt x=u$. If I've done all the algebra correctly, this leads to the equivalent inequality to prove,

$$1+3u^2e^{u^2}+3u\int_0^ue^{t^2}dt\le 3u^2+e^{3u^2}$$

From here you can use the Taylor series for the exponential function. The left-hand side becomes

$$1+\left(3u^2+3u^4+{3\over2}u^6+\cdots\right)+3u\left(u+{1\over3}u^3+{1\over10}u^5+\cdots\right)$$

while the right-hand side gives

$$3u^2+\left(1+3u^2+{9\over2}u^4+\cdots \right)$$

Note that each side, once terms are collected begins with $1+6u^2$. So all that remains is to look at the Taylor expansions closely enough to see that coefficients on the right (which all have powers of $3$ in their numerators) are individually greater than the corresponding coefficients on the left.

Answer 2

This is not a complete answer, but a start. We can rearrange the expression:

$$-2+e^x+\frac{1}{\sqrt{x}} \int^{\sqrt{x}}_0 e^{t^2} dt \leq \sum^{\infty}_{k=1} \frac{(3x)^{k-1}}{k!}-1$$

This is what I finally got:

$$\sum^{\infty}_{k=1} \frac{x^{k}}{k!}+\sum^{\infty}_{k=1} \frac{x^{k}}{(2k+1)k!} - \sum^{\infty}_{k=1} \frac{(3x)^{k-1}}{k!}\leq0$$

If I didn't make a mistake somewhere, then I think this is self-evident.