I have the following function: $$f(x,y)=\frac{xy}{1+x+y}$$ where $x\geq 0, y\geq0$. The Hessian of the above function is given as follows: $$\left[\matrix{-\frac{2y(y+1)}{(1+x+y)^3} & \frac{1+x+y+2xy}{(1+x+y)^3}\\ \frac{1+x+y+2xy}{(1+x+y)^3} & -\frac{2x(x+1)}{(1+x+y)^3}}\right]$$ I think its determinant is not always greater than zero so how to prove that the above function is concave? Thanks in advance.
I read about this function in a research paper. In that paper they say that $f(x,y)$ is a concave function because $-2y(y+1)(1+x+y)^2<0$ and determinant of the Hessian matrix is $\geq 0$ (In that paper the (second column and first row) and (first column and second row) entries of the Hessian matrix are written as $\frac{2xy}{(1+x+y)^3}$ instead of $\frac{1+x+y+2xy}{(1+x+y)^3}$) therefore I am confused.
I have even tried checking $$[v_1\quad v_2] \left[\matrix{-\frac{2y(y+1)}{(1+x+y)^3} & \frac{1+x+y+2xy}{(1+x+y)^3}\\ \frac{1+x+y+2xy}{(1+x+y)^3} & -\frac{2x(x+1)}{(1+x+y)^3}}\right]\Bigg[\matrix{v_1 \\v_2}\Bigg]$$ test for any real values of $v_1$ and $v_2$. The answer of the above matrix multiplication results in (ignoring the denominator since it is always positive) $$-y(y+1)v_1^2+(1+x+y+2xy)v_1v_2-x(1+x)v_2^2$$ I am not sure if it is always negative. Any help will be much appreciated.
The function is not concave.
On the line $(t,t)$, with $t \ge 0$, we have $\phi(t)=f(t,t) = {2t^2 \over 1+2t}$, and a bit of work shows that $\phi''(t) = {2 \over 8 t^3+ 12 t^2 +6 t +1} > 0$, so, in fact, $f$ is strictly convex on this line.