Am I correct to say that the following function is convex?
$\begin{align} & f(x,y)=-\sqrt{xy} \\ & x>0,y>0 \\ \end{align}$
After computing the Hessian:
$ Hf =\left[ \begin {array}{cc} 1/4\,{\frac {{y}^{2}}{ \left( xy \right) ^{ 3/2}}}&1/4\,{\frac {xy}{ \left( xy \right) ^{3/2}}}-1/2\,{\frac {1}{ \sqrt {xy}}}\\ 1/4\,{\frac {xy}{ \left( xy \right) ^ {3/2}}}-1/2\,{\frac {1}{\sqrt {xy}}}&1/4\,{\frac {{x}^{2}}{ \left( xy \right) ^{3/2}}}\end {array} \right]$
Which simplifies to:
$Hf=\left[ \begin {array}{cc} 1/4\,{\frac {y}{x\sqrt {xy}}}&-1/4\,{\frac {1}{\sqrt {xy}}}\\ -1/4\,{\frac {1}{\sqrt {xy}}}&1/4 \,{\frac {x}{y\sqrt {xy}}}\end {array} \right]$
And taking the determinant:
$det(Hf)=0 \ \ \ \ \forall \ x,y \in \Re^+$
Which is inconclusive.
Will need another method, namely $ z^T (H f) z \ $
See solution in the answer below for continuation:
Aside: And, extending this to an n-dimensional problem: $f(x_1,x_2,...,x_n)=-\sqrt[n]{x_1x_2...x_n}$ $x_i \gt 0 \ \ \ i=1,2,3,...,n$
Will also yield a convex function.
Thanks.