I am proving that $g(x,y)=\log(f(x,y))$ is a subharmonic function on $G$, if $f:G\to\mathbb{R}$ is subharmonic in $G$ (open set of $\mathbb{R}^{2}$) with $f(x,y)>0$ for all $(x,y)\in\mathbb{R}^{2}$. For that, I want to prove that the laplacian $$\Delta g=\frac{\partial^{2} g}{\partial x^{2}}+\frac{\partial^{2} g}{\partial y^{2}},$$ is greater or equal to zero in $G$. So, I start by computing the partial derivatives: $$ \frac{\partial^{2} g(x,y)}{\partial x^{2}}=\frac{\partial}{\partial x}\left(\frac{\frac{\partial f}{\partial x}(x,y)}{f(x,y)}\right)=\frac{\frac{\partial^{2} f}{\partial x^{2}}f(x,y)-\left(\frac{\partial f}{\partial x}(x,y)\right)^{2}}{(f(x,y))^{2}}. $$ $$ \frac{\partial^{2} g(x,y)}{\partial y^{2}}=\frac{\partial}{\partial y}\left(\frac{\frac{\partial f}{\partial y}(x,y)}{f(x,y)}\right)=\frac{\frac{\partial^{2} f}{\partial y^{2}}f(x,y)-\left(\frac{\partial f}{\partial y}(x,y)\right)^{2}}{(f(x,y))^{2}}. $$ Then, in $G$, $$\Delta g=\frac{\partial^{2} g}{\partial x^{2}}+\frac{\partial^{2} g}{\partial y^{2}}=\frac{\Delta f\cdot f-\left[\left(\frac{\partial f}{\partial x}\right)^{2}+\left(\frac{\partial f}{\partial y}\right)^{2}\right]}{f^{2}}.$$ Taking on account my hypothesis ($\Delta f\geq 0$, $f>0$), for seeing that $\Delta g\geq 0$, I just have to prove that $$ \Delta f\cdot f\geq\left(\frac{\partial f}{\partial x}\right)^{2}+\left(\frac{\partial f}{\partial y}\right)^{2}. $$ And this last step is where I am stuck... Is this inequality true for all $(x,y)$?
Thanks in advanced!