Is $f+g$ and $fg$ maximum where $f$ and $g$ are maximum? If $g$ and $f$ are two real functions and I know that both $f$ and $g$ have a maximum in $x_0 \in \mathbb{R}$ can I say that $f+g$ has a maximum in $x_0$?
If this is true, can the same be said if I have $fg$? That is, if  both $f$ and $g$ have a maximum in $x_0 \in \mathbb{R}$ can I say that $fg$ has a maximum in $x_0$?
 A: For the first question, yes: because $\forall x, f(x) \leq f(x_0)$ and $g(x) \leq g(x_0)$. So $(f+g)(x) \leq (f+g)(x_0)$.
For the second question: no. Let $f(x)=g(x)=1-x^2$. $f$ and $g$ have maximum in $0$. But $(fg)(2)=3 \times 3=9 > (fg)(0)=1$ 
A: This is true for $f+g$. Because for every $x$ in the neighborhood of $x_0$ you have
\begin{aligned}
f(x)&\leq f(x_0)\\g(x)&\leq g(x_0)
\end{aligned}
So, by adding the above inequalities you'll get $$f(x)+g(x)\leq f(x_0)+g(x_0)$$ or $$(f+g)(x)\leq (f+g)(x_0)$$ which means the maximum of $f+g$ occurs at $x_0$, too.
But the proposition is not true for $f\cdot g$. For example you can consider both of $f$ and $g$ to be negative functions, like $$f(x)=g(x)=-(1+x^2)$$ for which the maximum occurs at $x_0=0$ . You can see $$(f\cdot g)(x)=\left(1+x^2\right)^2$$ which its minimum occurs at $x_0=0$ and also has no maximum.
A: Assuming that $f$ and $g$ are twice differentiable, we can classify critical points by the second derivative test. For the first part, we have $(f + g)' = f' + g'$ and so $x_{0}$ is a critical point of $f + g$. Moreover, if $x_{0}$ if a maximum of $f$ and $g$, what can you say about the sign of $f''$ and $g''$ and how could this help determine the sign of $(f + g)''$? 
Now $(fg)' = fg' + f'g$, so again $x_{0}$ is a critical point of $fg$. But $(fg)'' = (fg' + f'g)' = fg'' + 2f'g' + f''g$, and it becomes hard to say whether or not $x_{0}$ is a maximum. As marco has pointed out in another answer, there are counter examples. 
A: Not necesserly, if $f(x_0)<0$ and $g(x_0)<0$ then it is a minimum of $fg$
