Convexity of ($x^{1/4}+y^{1/4}+z^{1/4})^3$ I´m asked to determine if this function is convex, concave or neither.
($x^{1/4}+y^{1/4}+z^{1/4})^3$ for $x>0$, $y>0, z> 0$
The inside function is a sum of concave functions so is concave, the outside function is $x^3$ which is convex for $x>0$ and is increasing too.
With this in mind, can I conclude that is not either convex or concave? Or there is a relationship for that situation?
Thanks.
 A: Yes, you can conclude that $f$ is convex. Define $g :\mathbb R_{++}^3 \rightarrow \mathbb R_{++}$ as the function $(x,y,z) \mapsto x^{1/4} +y^{1/4}+z^{1/4}$ and $h: \mathbb R_{++} \rightarrow \mathbb R$ as the function $x \mapsto x^3$. ($\mathbb R_{++}^k$ denotes the vector with all positive entries.)
Then as shown in Boyd and Vanderberghe's Convex Optimization, in Section 3.2.4 of Chapter 3, but the summary of the argument is as follows:
Application of the chain rule (twice) shows that
$$\nabla f (v) = \nabla g (v) \nabla h(g(v))$$
$$ \Rightarrow \nabla^2 f(v) = h''(g(v)) \big(\nabla g (v)\big)^T \nabla g(v) + h'(g(v)) \nabla^2 g(v) $$
Now recall that a twice differentiable function is convex on its domain if and only if the second derivative is positive semidefinite on the entire domain. The fact that $h$ is nondecreasing and convex means $h'' > 0$ and $h' > 0$.
From the above we can conclude $\nabla^2 f$ is positive semidefinite on the domain we defined it on because it is the sum of positive scalings of the positive semidefinite matrices $\nabla g(v)^T \nabla g(v)$ and $\nabla^2 g(v)$, where the former is psd by definition, and the latter is posiitve semidefinite since $g$ is convex because of how we defined it.
