Wolfram alpha error in global minimization? Let us consider the function 
$$f(x,y)= x + y^2 - \ln(x+y)$$
If you try to minimize it using Wolfram Alpha (http://www.wolframalpha.com/input/?i=minimize+x%2By%5E2-ln%28x%2By%29), it founds a local minimum at $(x,y)=(1/2,1/2)$, but it can't find a global minimum. 
But that local minimum should also be the global one, since the function is a convex one and the domain is convex.
Am I wrong or is Wolfram Alpha wrong?

Ok, thanks to everybody who answered or commented, but perhaps I was not clear when writing my question: I knew that $(1/2,1/2)$ is a local minimum, and, since the $f(x,y)=x+y^2-ln(x+y)$ is a convex function and the domain is $D=\{(x,y) \in \mathbb{R}^2 : x+y >0\}$, which is a convex subset of $\mathbb{R}^2$, then a well kwown result in convex optimization ensures that $(1/2,1/2)$ is also a global minimum.
Then perhaps I should have written: "why does Wolfram Alpha says 'no global minima found' when I ask to minimize $f(x,y)$?"
As far as I can understand from your answers, that sentence from Wolfram Alpha means only that it is not able to find a global minimum, not that the global minimum does not exist, is this correct?
 A: The domain of $f$ is the set of pairs $(x,y)$ for which $r=x+y>0.$ Using $x=r-y$ our function can be expressed nicely in terms of separate functions of $r,y$ as 
$$(r-\ln r)+(y^2-y).$$
Using one variable calc, the min of the first bracketed term is $1$ (at $r=1$), while the min of the second is $-1/4$ (at $y=1/2$).
This shows your function satisfies $f \ge 1 -1/4=3/4.$ That this is actually the global minimum then follows since $f(1/2,1/2)=3/4.$
If Wolfram explicitly says no global minimum exists, that is incorrect. However in some cases a computer algebra system may not deal with implied domains well... [I just looked at Wolfram output you link to, and it only says there is a local min, not that there is not a global one.]
A: Find the partial derivatives:
$$\frac{∂}{∂x}(f(x,y))=1-\frac{1}{x+y}\:\:\:\:(1)$$
and
$$\frac{∂}{∂y}(f(x,y))=2y-\frac{1}{x+y}\:\:\:\:(2)$$
Set $(1)$ equal to $0$:
$$0=1-\frac{1}{x+y}$$
$$\frac{1}{x+y}=1$$
$$x+y=1\:\:\:\:(3)$$
Set $(2)$ equal to $0$:
$$0=2y-\frac{1}{x+y}\:\:\:\:(4)$$
Put $(3)$ into $(4)$:
$$0=2y-\frac{1}{1}$$
$$y=\frac{1}{2}\:\:\:\:(5)$$
Put $(5)$ back into $(3)$:
$$x+\frac{1}{2}=1$$
$$x=\frac{1}{2}$$
So $(x,y) = (\frac{1}{2},\frac{1}{2})$ is our potential extremum. Now find our second partial derivatives, and perform the second derivative test for a function of two variables:
$$f_{xx} = \frac{1}{(x+y)^2}$$
$$f_{xy} = f_{yx}=\frac{1}{(x+y)^2}$$
$$f_{yy}= \frac{1}{(x+y)^2} + 2$$
And now the test:
$$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2$$
$$D(\frac{1}{2},\frac{1}{2})=(\frac{1}{(\frac{1}{2}+\frac{1}{2})^2})(\frac{1}{(\frac{1}{2}+\frac{1}{2})^2} + 2)+(\frac{1}{(\frac{1}{2}+\frac{1}{2})^2})^2$$
$$D(\frac{1}{2},\frac{1}{2})=4$$
Also note that:
$$f_{xx}(\frac{1}{2},\frac{1}{2})=\frac{1}{(\frac{1}{2}+\frac{1}{2})^2}=1$$
And so, we have $D(\frac{1}{2},\frac{1}{2})>0$ and $f_{xx}(\frac{1}{2},\frac{1}{2})>0$, so by the second derivative test for a function of two variables, we conclude that $(\frac{1}{2}, \frac{1}{2})$ is a local minimum, and the only extremum. As you stated in your question, this point must be a global minimum. WolframAlpha simply said it was a local minimum, and didn't say that it wasn't a global one:

A: Let's look at the first order conditions: 
$f_{x} = 1 - \frac{1}{x + y} = 0$ 
$f_{y} = 2y - \frac{1}{x + y} = 0$ 
So $\frac{1}{x + y} = 1$, which gives us $y = \frac{1}{2}$. It follows that $x = \frac{1}{2}$. 
So Wolfram Alpha seems to be correct here. Note you should also verify that $(\frac{1}{2}, \frac{1}{2})$ is a minimum, but this is straight-forward to do.
