Is the utility function $u(x)=x_1 x_2 + \gamma x_2$ concave or quasi-concave?

Question

I want to prove, given $\gamma>0$ and $x\in \mathbb{R}^2_+$, if the utility function:

$$u(x)=x_1 x_2 + \gamma x_2$$

is concave, strictly concave, quasi-concave or strictly quasi-concave.

I have tried with Hessian but I do not reach anywhere.

Then I tried with the definition of concavity and quasi-concavity and I got that it is strictly quasi-concave if (supossing without loss of generality that $u(y)>u(x)$ ): $$\alpha(y_1-x_2)(y_2-x_2) < (y_1 y_2 + \gamma y_2)(x_1 x_2 + \gamma x_2)$$ Which I think is sattisfied.

Can someone give me a hint on this?

Answer 1

Consider $A_\alpha = \{(x_1, x_2) \in \mathbb{R}_+^2: (x_1 + \gamma)x_2 \ge \alpha\}$.

If $\alpha \le 0$, then $A_\alpha = \mathbb{R}_+^2$ which is convex but not strictly convex.

If $\alpha > 0$, then $A_{\alpha} = \{(x_1, x_2)\in \mathbb{R}_+^2:x_2 \ge \frac{\alpha}{x_1+\gamma}\}$ which is convex.

Hence $u$ is quasi-concave.

It is easay to check that it is not concave by examining the Hessian.

Answer 2

The hessian matrix is \begin{equation} H= \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \end{equation}

Therefore \begin{equation}x^THx=\begin{bmatrix} x_1\\ x_2 \end{bmatrix}^T\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix}=2x_1x_2\geq 0 \end{equation}