Prove the convexity for f: $\mathbb{R}^3$ $\to$ $\mathbb{R}$

Question

The problem below is one of my homework for the Convex Geometry course. I have tried to prove by using the Hessian matrix or the gradient but cannot figure out the right answers.

Show that $$ f(x_1, x_2, x_3) = \cfrac1{x_1-\cfrac{1}{x_2-\cfrac1{x_3}}} $$ is convex on $\left(\Bbb R_*^+\right)^3$.

Can you help me on this, please?

Answer 1

This function is not convex. Set $\alpha = 1/2$. Then

\begin{align} 0.75 =f(2,2,2) &= f(\alpha(1,1,1) + (1-\alpha)(3,3,3)) \\ &> \alpha f(1,1,1) + (1-\alpha)f(3,3,3) = \alpha 0 + (1-\alpha)\frac{15}{56} \\ &\approx 0.134 \end{align}

which violates the convexity inequality. You can see this from just looking at the subset of $f$'s graph where $x_1=x_2=x_3$.