Prove that the function $f(x_1 , x_2) = x_1 x_2$ is quasi-concave on $S = \mathbb{R}_{++}^2$
I started with the definition of quasi-concavity:
$$ f( \lambda x + (1-\lambda) y) \geq \min \left( f(x), f(y) \right) $$
Case 1:
$$(λ * x_1 + (1-λ) * y_1) * (λ * x_2 + (1-λ) * y_2) ≥ x_1 * x_2$$
Case 2:
$$(λ * x_1 + (1-λ) * y_1) * (λ * x_2 + (1-λ) * y_2) ≥ y_1 * y_2$$
But calculating it for the first case I get:
$$x_1 * x_2 * (λ^2-1) + (1-λ) * λ * y_1 * x_2 + (1-λ) * λ * x_1 * y_2 + (1-λ)^2 * y_1 * y_2 ≥ 0$$
And I think now I have to proof that for all positive $x_1,x_2,y_1,y_2$ the equation is satisfied. But is that even possible? I think it depends on how one chooses λ.
Best Answer
Let $(x_1, x_2) \in S$ and $(y_1, y_2) \in S$ and $$ c = \min \{ f(x_1, x_2), f(y_1, y_2) \} = \min \{ x_1 x_2, y_1 y_2 \} \, . $$
For $0 \le \lambda \le 1$ is $$ f(\lambda x_1 + (1-\lambda) y_1, \lambda x_2 + (1-\lambda) y_2) = (\lambda x_1 + (1-\lambda) y_1)(\lambda x_2 + (1-\lambda) y_2) \\ = \lambda^2 x_1 x_2 + \lambda (1-\lambda)(x_1 y_2 + x_2 y_1) + (1-\lambda)^2 y_1 y_2 \, . $$ Now $x_1 x_2 \ge c$, $y_1 y_2 \ge c$, and using the inequality between arithmetic and geometric mean, $$ x_1 y_2 + x_2 y_1 \ge 2 \sqrt{x_1 y_2 x_2 y_1} \ge 2c \, . $$ It follows that $$ f(\lambda x_1 + (1-\lambda) y_1, \lambda x_2 + (1-\lambda) y_2) \ge c \bigl( \lambda^2 + 2\lambda(1-\lambda) + (1-\lambda)^2\bigr) = c $$ and that concludes the proof.
Alternatively one can argue that for each $c > 0$ the sublevel set $$ \{ (x_1, x_2) \in S \mid x_1x_2 \ge c \} = \{ (x,y) \mid x > 0, y \ge \frac c x\} $$ is convex as the epigraph of the convex function $x \mapsto c/x$ with the domain $(0, \infty)$.