# Proof of the inequality $e^x \leq x+e^{\frac{9}{16}x^2}$

How can we prove the inequality $$e^x \leq x+e^{\frac{9}{16}x^2}$$?

Furthermore, if $$e^x \leq x+e^{Cx^2}$$ holds, what is the smallest $$C$$?

It is similar to the question proof of inequality $e^x\le x+e^{x^2}$, and I have tried all the approaches mentioned in it, but the bound here is sharper, and the problem is more difficult.

For example, we need to prove that $$e^{\frac{9}{16}x^2-x}+x e^{-x}\geq 1$$, but when I get for $$x\geq 0$$, $$e^{\frac{9}{16}x^2-x}+x e^{-x}\geq 1+\frac{9}{16}x^2-x + x(1-x)=1-\frac{7}{16}x^2,$$ I can't go ahead any further.

• You can get a lower bound for $C$ if you consider $x\rightarrow 0^+$ for $$\frac{e^x-x-1}{x^2} \leq \frac{e^{Cx^2}-1}{x^2}.$$ The Taylor series allows you to evaluate the limits rather quickly. Commented May 4, 2022 at 3:35
• Taking logarithm and one can see that the smallest $C$ should be the maximum of $\frac{\log(e^x-x)}{x^2}$, or about $0.5574$. I don't think that it has a closed form, though. Commented May 4, 2022 at 3:50
• @JianingSong. I was typing the same ! Commented May 4, 2022 at 3:55

For $$x\le 0$$, let $$g(x) = e^\frac{x^2}{2}-e^x+x$$, then $$g'(x) = xe^\frac{x^2}{2}-e^x+1$$, $$g''(x) = (x^2+1)e^\frac{x^2}{2}-e^x\ge (x^2+1)e^\frac{x^2}{2}-1\ge 0$$. So $$g'(x)$$ increases on $$(-\infty,0]$$, hence $$g'(x)\le g'(0)=0$$ for all $$x\le 0$$; $$g(x)$$ decreases on $$(-\infty,0]$$, so $$g(x)\ge g(0)=0$$ for all $$x\le 0$$. This shows that $$e^x\le x+e^\frac{x^2}{2}$$ for $$x\le 0$$.
Now suppose that $$x\ge 0$$. Note that $$e^x\ge 1+x+\dfrac{x^2}{2!}+\cdots+\dfrac{x^n}{n!}$$ for all odd $$n$$ and $$x\in\mathbb{R}$$, so we have $$e^{\frac{9}{16}x^2-x} + xe^{-x}-1\ge 1+\left(\frac{9}{16}x^2-x\right)+\frac{1}{2}\left(\frac{9}{16}x^2-x\right)^2+\frac{1}{6}\left(\frac{9}{16}x^2-x\right)^3+x\left(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}-\frac{x^5}{120}\right)-1,$$ which simplifies to $$e^{\frac{9}{16}x^2-x} + xe^{-x}-1\ge \frac{x^2}{122880}(2621x^4 - 14320x^3 + 33520x^2 - 28160x + 7680).$$ It remains to show that $$2621x^4 - 14320x^3 + 33520x^2 - 28160x + 7680\ge 0$$. Well, there are several ways to prove this, for example by noting that $$2621x^4 - 14320x^3 + 33520x^2 - 28160x + 7680 = \frac{1}{10484}(5242x^2-14320x+7049)^2 + \frac{3}{15826376437}(6038297x-3889420)^2 + \frac{1763413283}{24153188}.$$