I cannot figure out how to show this function is convex. When I twice differentiate it the hessian does not have positive diagonal values and it shows as concave. I know it has both a concave and convex part of the graph as I have plotted, so how do I show the convex part?
$$f(x,y) = \frac{xy}{x+y}$$
Hint (without Hessian matrix otherwise see Convexity, Hessian matrix, and positive semidefinite matrix).
Note that for $t\in (0,1)$, and for $(x_1,y_1),(x_2,y_2)\in D_+:=\{(x,y): x+y>0\}$ (which is a convex set), $$f(tx_1+(1-t)x_2,ty_1+(1-t)y_2)-tf(x_1,y_1)-(1-t)f(x_2,y_2)\\= \frac{t(1-t)(x_1y_2-x_2y_1)^2}{(x_1+y_1)(x_2+y_2)(tx_1+(1-t)x_2+ty_1+(1-t)y_2)}\geq 0.$$ What about $D_-:=\{(x,y): x+y<0\}$? What may we conclude?
