Prove that the function is convex Question is reopened due to the problem found in the original solution.

I have the following function:
$$ A(v) = -\dfrac{k-1}{\displaystyle\sum_{i=1}^k \frac{1}{v_i}}, $$
where $k\geq 2$ is some integer constant and $1 \leq v_i \leq k-1$.
I am trying to prove that the function $A(v)$ is convex. According to Wolfram's FunctionConvexity, it is. Therefore, the function is somehow passes the following condition:
$$ f(t x +(1-t)y)\leq t f(x)+(1-t)f(y) ,$$
where $0\leq t\leq 1.$
However, when I have tried to prove this inequality myself I have bumped into a very messy inequality, which looks intractable to me. I have also tried to prove the positive semidefiniteness of the Hessian matrix following Sylvester's criterion, but it seems that the last leading principal minor equals zero. How could I approach this problem?

Update 1: I reopen the question due to the problem found in the solution. The proposed solution is possibly wrong. $h$ is convex decreasing on positive reals, but $g$ assumes negative real values. I am not sure how to resolve it.
Update 2: Here you can find additional information on how FunctionConvexity works. It was well explained in the first answer.
 A: The solution follows the suggestion by Rodrigo de Azevedo.
Firstly, observe that
$$ A(v) = \dfrac{k-1}{-\displaystyle\sum_{i=1}^k \frac{1}{v_i}},$$
is composition $A(v)=h \circ g=h(g(v))$ where:
$$g(v)= -\sum_{i=1}^k \frac{1}{v_i}, $$
and
$$h(x)=\frac{k-1}{x}.$$
Now observe that $h(x)$ is a convex decreasing function on the positive reals. Moreover, its extended-value extension $\tilde{h}(x)$ is nonincreasing. Given that g(v) is concave for any $1 \leq v_i\leq k-1$, it then follows that $A(v)$ is convex.
The conclusion follows from the statement that can be found on page 84 of Boyd & Vandenberghe.


A: Let
$$g(v) = \frac{1}{\sum_{i=1}^k \frac{1}{v_i}}.$$
It is known that $g(v)$ is concave on $\mathbb{R}_{> 0}^k$.
This result is given in Page 116, 3.17, [1].
In general, suppose $p < 1, p\ne 0$, the function
$$f(x) = \left(\sum_{i=1}^n x_i^p\right)^{1/p}$$
is concave on $\mathbb{R}_{> 0}^n$.
The proof is given in the solutions manual of the book [1]. (I put that proof at the end.)
Reference:
[1] Boyd and Vandenberghe, "Convex optimization".
http://web.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf

Proof of concavity of $f(x)$:
The first derivatives of $f$ are given by
$$\frac{\partial f(x)}{\partial x_i} = 
\left(\sum_{i=1}^n x_i^p\right)^{1/p - 1}x_i^{p - 1} = \left(\frac{f(x)}{x_i}\right)^{1 - p}, \quad \forall i.$$
The second derivatives are
$$\frac{\partial^2 f(x)}{\partial x_i \partial x_j} = \frac{1 - p}{x_i}\left(\frac{f(x)}{x_i}\right)^{ - p} \left(\frac{f(x)}{x_j}\right)^{1 - p} = \frac{1 - p}{f(x)}\left(\frac{f(x)^2}{x_ix_j}\right)^{1 - p}, \quad i\ne j,$$
and
$$\frac{\partial^2 f(x)}{\partial x_i^2} = \frac{1 - p}{f(x)}\left(\frac{f(x)^2}{x_i^2}\right)^{1 - p} - \frac{1 - p}{x_i}\left(\frac{f(x)}{x_i}\right)^{1 - p}.$$
It suffices to prove that, for all $y \in \mathbb{R}^n$,
$$y^\mathsf{T}\nabla^2 f(x) y
= \frac{1 - p}{f(x)}
\left[\left(\sum_{i=1}^n \frac{y_i f(x)^{1 - p}}{x_i^{1 - p}}\right)^2 - \sum_{i=1}^n \frac{y_i^2f(x)^{2 - p}}{x_i^{2 - p}}\right] \le 0.$$
This follows by applying the Cauchy-Bunyakovsky-Schwarz inequality
$(a^\mathsf{T}b)^2 \le \|a\|_2^2\|b\|_2^2$ with
$$a_i = \left(\frac{f(x)}{x_i}\right)^{-p/2},
\quad b_i = y_i \left(\frac{f(x)}{x_i}\right)^{1 - p/2},$$
and noting that $\sum_{i=1}^n a_i^2 = 1$.
We are done.
A: Let $(x_1, \ldots, x_n) \in \mathbb R_{>0}^n$ and $(v_1, \ldots, v_n) \in \mathbb R_{>0}^n$. By Schwarz’ inequality $$\begin{eqnarray}
\frac1{x_1}+ \ldots + \frac1{x_n} &=& \left(\frac{\sqrt{v_1}}{x_1}, \ldots, \frac{\sqrt{v_n}}{x_n} \right) \cdot \left(\frac1{\sqrt{v_1}}, \ldots, \frac1{\sqrt{v_n}} \right) \\
 & \leq & \left(\frac{v_1}{x_1^2} + \ldots + \frac{v_n}{x_n^2} \right)^{\frac12} \left(\frac1{v_1} + \ldots + \frac1{v_n} \right)^{\frac12}.
\end{eqnarray}$$
Rewrite this as
$$\left(\frac1{v_1} + \ldots + \frac1{v_n}\right)^{-1} \leq \left(\frac{v_1}{x_1^2} + \ldots + \frac{v_n}{x_n^2}\right) \left(\frac1{x_1} + \ldots + \frac1{x_n}\right)^{-2}.$$
Now the right hand side is exactly the tangent plane of the left hand side at the point $(x_1, \ldots, x_n)$. Since this holds for any such point $(x_1, \ldots, x_n)$ the left hand side is a concave function.
A: Consider the Hessian matrix of $D(v)$:
$$
H=
\begin{bmatrix}
 \frac{\partial^2 D(v)}{\partial v_1^2} & \frac{\partial^2 D(v)}{\partial v_1 \partial v_2}& \dots & \frac{\partial^2 D(v)}{\partial v_1 \partial v_k} \\
\frac{\partial^2 D(v)}{\partial v_2 \partial v_1}&\frac{\partial^2 D(v)}{\partial v_2^2}  & \dots & \frac{\partial^2 D(v)}{\partial v_2 \partial v_k} \\
\vdots  & \vdots & \ddots & \vdots\\
 \frac{\partial^2 D(v)}{\partial v_k \partial v_1}  & \frac{\partial^2 D(v)}{\partial v_k \partial v_2}  & \dots &  \frac{\partial^2 D(v)}{\partial v_k^2}
\end{bmatrix}
$$
Observe that the diagonal of the matrix consists of positive elements of the form:
$$\frac{ \partial^2 D(v)}{\partial v_i^2} = \frac{2 \left(k-1 \right)  \prod^k_{j\neq i}v_i^2  \left( \sum_{j \in (1,k]} \prod_{p\neq j} v_p \right) }{\left( \sum^k_{j=1} \prod_{i\neq j}^k v_i \right)^3 }.$$
Off-diagonal elements are represented by negative mixed partial derivatives of the form:
$$ \frac{ \partial^2 D(v)}{\partial v_i \partial v_j} =-\frac{2 \left(k-1 \right) v_i^2 v_j^2  \prod_{p=1}^k v_i }{\left( \sum^k_{j=1} \prod_{i\neq j}^k v_i \right)^3 }. $$
Now, the first leading principal minor is:
$$   \Delta_1 =   \frac{2 \left(k-1 \right)  \prod_{j\neq 1}^k v_j^2  \left( \sum_{j \in (1,k]} \prod_{p\neq j} v_p \right) }{\left( \sum^k_{j=1} \prod_{i\neq j}^k v_i \right)^3 } > 0.  $$
The second leading principal is:
$$ \Delta_2 =\Delta_1  \frac{2(k-1)\prod_{j\neq 2}^k v_j^2  \left( \sum_{j \in (2,k]} \prod_{p\neq j} v_p \right)}{ \left( \sum^k_{j=1} \prod_{i\neq j}^k v_j \right)^2 \left( \sum_{j \in (1,k]} \prod_{p\neq j} v_p \right) } >0 .$$
Assume that $m<k$ leading principal minor $\Delta_m>0$. Now I verify that $\Delta_{m+1}$ is also positive:
\begin{equation}
\label{m plus one}
 \Delta_{m+1} = \Delta_m   \frac{2(k-1)\prod_{j\neq m}^k v_j^2  \left(  \overbrace{\sum_{j \in (m+1,k]} \prod_{p\neq j} v_p }^{n_1}\right)}{ \left( \sum^k_{j=1} \prod_{i\neq j}^k v_j \right)^2 \left( \sum_{j \in (m,k]} \prod_{p\neq j} v_p \right) }>0, 
\end{equation}
which is always positive too.
Therefore, the first $m<k$ leading principal minors are positive by induction.
Also observe that term in $\Delta_{m+1}$ term $ n_1=0$ if $m+1=k$ and the last leading principal minor is $\Delta_k=0$.
Given that matrix, $H$ enjoys leading implies all (LIA) properties and all leading principal minors are positive, it immediately follows that $H$ is positive semi-definite and $D(v)$ is convex.
