Given a scalar field, how to show that the eigenvalues of its Hessian are bounded

Question

Show that the eigenvalues of the Hessian of $$f(x_1,x_2) := x_1^2+x_1x_2+x_2^2+\ln (1+2e^{x_2})$$ are bounded, i.e., $$1 \leq \lambda_{\min} \left(\nabla^2 f(x)\right) \leq \lambda_{\max}\left(\nabla^2 f(x)\right) \leq 4$$

What I tried is finding gradient and then Hessian

$$
\nabla f(x) =
\begin{bmatrix}
2x_1+x_2 \\
x_1+2x_2+\frac{2e^{x_2}}{1+2e^{x_2}}
\end{bmatrix}
$$

and Hessian is:

$$
\nabla^2 f(x) =
\begin{bmatrix}
2 & 1 \\
1 & 2
+\frac{2e^{x_2}}{(1+2e^{x_2})^2}
\end{bmatrix}
$$

Let $y=\frac{2e^{x_2}}{(1+2e^{x_2})^2}$. Then

$$
\nabla^2 f(x) =
\begin{bmatrix}
2 & 1 \\
1 & 2
+y
\end{bmatrix}
$$

Then,

$$
\lambda_{1,2} = \frac{4+y\pm \sqrt{y^2+4}}{2}
$$

My questions are:

1. Am I mistaken?

2. Is this function a well-known function?

Answer 1

Hint: Show that $y$ is bounded for all $x_2\in\mathbb{R}$:

Part 1:

\begin{align}y=\frac{2e^{x_2}}{(1+2e^{x_2})^2}< \frac{1+2e^{x_2}}{(1+2e^{x_2})^2}=\frac{1}{1+2e^{x_2}}<1\end{align}

Part 2: Observe that \begin{align} \frac{2e^{x_2}}{(1+2e^{x_2})^2}=\frac{-1+1+2e^{x_2}}{(1+2e^{x_2})^2}=-\frac{1}{(1+2e^{x_2})^2}+\frac{1}{1+2e^{x_2}}\end{align} but since $(1+2e^{x_2})^2>1+2e^{x_2}$ we have $$\frac{1}{1+2e^{x_2}}>\frac{1}{(1+2e^{x_2})^2}$$ and thus $$-\frac{1}{(1+2e^{x_2})^2}+\frac{1}{1+2e^{x_2}}>0$$ Consequently

\begin{align}y=\frac{2e^{x_2}}{(1+2e^{x_2})^2}=-\frac{1}{(1+2e^{x_2})^2}+\frac{1}{1+2e^{x_2}}>0 \end{align}

Now you can do the final part with $0<y<1$