Does anybody have a good way to prove the inequality $e^x\leq x+e^{9x^2/16}$ ?
I found this here
Wainwright, M. J. (2019). High-dimensional statistics: A non-asymptotic viewpoint (Vol. 48). Cambridge University Press.
On page 47, the author directly used this
inequality
By the way, here are some good proofs of $e^x\leq x+e^{x^2}$.
proof of inequality $e^x\le x+e^{x^2}$
I have to mention that the first-order derivative of this function $f(x)=x+e^{9x^2/16}-e^x$ has multiple roots as below.
The first-order derivative: $f'(x)=\frac{9xe^{\frac{9x^2}{16}}}{8}+1-e^x$
Plot of the first-order derivative:
the first-order derivative plot
Three roots: 0,0.41490879,0.54479289
By the way, can you find the $\min$ value of $c$ such that the inequality $e^x\leq x+e^{cx^2}$ is satisfied? I tried several values of $c$ less than 9/16, and they work!
Best Answer
Remarks: The optimal $c$ is given by $c_0 = \max_{x\ne 0} \frac{\ln(\mathrm{e}^x - x)}{x^2}$. I think that it does not admit a closed form. Here we give a proof of $c = 9/16$.
Letting $x = \frac43 y$, the desired inequality is written as $$\mathrm{e}^{y^2} + \frac43y \ge \mathrm{e}^{4y/3}. \tag{1}$$
Case 1: $y \ge 4/3$
Clearly (1) is true.
$\phantom{2}$
Case 2: $y < 4/3$
Using $\mathrm{e}^u \ge 1 + u + \frac12u^2 + \frac16u^3$ for all $u\ge 0$, we have \begin{align*} \mathrm{e}^{y^2} &\ge 1 + y^2 + \frac12y^4 + \frac16 y^6\\[6pt] &\ge 1 + y^2 + \frac12y^4 + \frac16 \cdot \left(\frac12 y^4 - \frac{1}{16}y^2\right) \tag{2}\\[6pt] &= \frac{7}{12}y^4 + \frac{95}{96}y^2 + 1\\[6pt] &\ge \frac{33}{35}\cdot \frac{7}{12}y^4 + \frac{95}{96}y^2 + 1\\[6pt] &= \frac{11}{20}y^4 + \frac{95}{96}y^2 + 1 \end{align*} where in (2) we have used $y^6 + \frac{1}{16}y^2 \ge \frac12 y^4$ (AM-GM).
Thus, it suffices to prove that $$\left(\frac{11}{20}y^4 + \frac{95}{96}y^2 + 1\right) + \frac43y \ge \mathrm{e}^{4y/3}.$$
Since $\frac{11}{20}y^4 + \frac{95}{96}y^2 + \frac43y + 1 > 0$, it suffices to prove that $$\ln\left(\frac{11}{20}y^4 + \frac{95}{96}y^2 + \frac43y + 1\right) \ge \frac{4y}{3}.$$ Denote $\mathrm{LHS} - \mathrm{RHS}$ by $f(y)$. We have $$f'(y) = \frac{-1056y(y - a)(y- 1/2)(y - b)}{792y^4 + 1425y^2 + 1920y + 1440}$$ where $$a = \frac54 - \frac{1}{132}\sqrt{17655} \approx 0.2434, \quad b = \frac54 + \frac{1}{132}\sqrt{17655} \approx 2.2566.$$ Thus, we have $f'(y) < 0$ on $(-\infty, 0)$, and $f'(y) > 0$ on $(0, a)$, and $f'(y) < 0$ on $(a, 1/2)$, and $f'(y) > 0$ on $(1/2, 4/3)$. Also, we have $f(0) = 0$ and $f(1/2) = \ln\frac{1247}{640} - \frac23 > 0$. Thus, $f(y)\ge 0$ for all $y < 4/3$.
We are done.