Convexity of the sublevel set of $g(a,b,c)=(a^{-1}+b^{-1}+c^{-1})(a+b+c)$ Consider the function $g(a,b,c) = (a^{-1}+b^{-1}+c^{-1})(a+b+c)$ defined in the domain $a,b,c >0$ and its sublevel set $g_{\alpha} = \{ (a,b,c) \in \mathbb{R}^3 | g(a,b,c)\leq \alpha\}$, I would like to prove that $g_{\alpha}$ is convex by definition.
I would take any $x_{1} = (a_1, b_1, c_1)$ and $x_{2} = (a_2, b_2, c_2) \in g_{\alpha}$, and would like to show that $\lambda x_{1} + (1-\lambda)x_{2} \in g_{\alpha}$
My method is pretty standard, with the goal of proving that $g(\lambda x_{1} + (1-\lambda)x_{2}) \leq \alpha$, expanding $g(\lambda x_{1} + (1-\lambda)x_{2})$ yields:
$$g(\lambda x_{1} + (1-\lambda)x_{2}) = (\frac{1}{\lambda a_1 + (1-\lambda) a_2} +\frac{1}{\lambda b_1 + (1-\lambda) b_2} + \frac{1}{\lambda c_1 + (1-\lambda) c_2})(\lambda a_1 + (1-\lambda) a_2 + \lambda b_1 + (1-\lambda) b_2 + \lambda c_1 + (1-\lambda) c_2$$
Simplifying yields:
$$g(\lambda x_{1} + (1-\lambda)x_{2}) = (\frac{1}{\lambda a_1 + (1-\lambda) a_2} +\frac{1}{\lambda b_1 + (1-\lambda) b_2} + \frac{1}{\lambda c_1 + (1-\lambda) c_2}) (\lambda(a_1 + b_1+c_1) + (1-\lambda)(a_2 + b_2 + c_2))$$
I am not sure on how to deal with the fraction term, but I suspect there should be a easier way than combining the fractions and trying to simplify it (I have tried it for the 2 variable case $g(a,b) = (a^{-1}+b^{-1})(a+b)$, but it became a mess which I am not sure on how to continue on to simplify.
Any help on this is greatly appreciated!
 A: (Note: I left out some calculations in some steps, just leave a comment if
I should include more calculations. I think the general idea is more useful
than the detailed calculations for the purpose of this answer.)
To simplify things, we will use the notation
$s_i:=a_i+b_i+c_i$ whenever variables $a_i>0,b_i>0,c_i>0$ appear.
First, one can show that $g(x_i)=g(t_i x_i)$
holds for all $t_i>0$.
In particular, we have $g(x_i)=g(s_i^{-1}x_i)$.
Therefore, it might be a good idea to consider the normalized points $s_i^{-1}x_i$.
It turns out that one can show the equality
$$
\begin{aligned}
 \frac{\lambda x_1+(1-\lambda) x_2}{\lambda s_1+(1-\lambda) s_2}
 & =
 \frac{\lambda s_1}{\lambda s_1+(1-\lambda) s_2}\cdot \frac{x_1}{s_1}
 + \frac{(1-\lambda) s_2}{\lambda s_1+(1-\lambda) s_2}\cdot \frac{x_2}{s_2}
 \\ & =
 \mu\frac{x_1}{s_1} + (1-\mu)\frac{x_2}{s_2},
\end{aligned}
$$
if we define $\mu:=\frac{\lambda s_1}{\lambda s_1+(1-\lambda) s_2}\in [0,1]$.
So we have some kind of convex combination, just with different weights.
It can also be shown that $g$ is convex on the set
$$E:=\{(a,b,c)\in\Bbb R^3 | a,b,c>0, a+b+c=1\}.$$
Indeed, for $(a,b,c)\in E$ we have
$g(a,b,c)=a^{-1}+b^{-1}+c^{-1}$, which is the sum of three convex functions
(on $(0,\infty)^3$) and therefore convex.
We also mention that $\frac{x_i}{s_i}\in E$ and $\mu\frac{x_1}{s_1} + (1-\mu)\frac{x_2}{s_2}\in E$ hold.
Then, combining all these observations we have
$$
\begin{aligned}
g(\lambda x_1+(1-\lambda x_2))
&= g(\frac{\lambda x_1+(1-\lambda x_2)}{\lambda s_1+(1-\lambda) s_2})
\\ &= 
g(\mu\frac{x_1}{s_1}+(1-\mu)\frac{x_2}{s_2})
\\ &\leq \mu g(\frac{x_1}{s_1})+(1-\mu)g(\frac{x_2}{s_2})
\\ &= \mu g(x_1)+(1-\mu)g(x_2)\leq \alpha.
\end{aligned}
$$
