If $p$ and $q$ are positive real numbers such that $p^2+q^2=1$, then the maximum value of $(p+q)$ is:

We know that $(p+q{)}^2=p^2+q^2+2pq$.Given $p^2+q^2=1$, we have $(p+q{)}^2=1+2pq$.By AM-GM inequality for $p^2$ and $q^2$:
$$\frac{p^2+q^2}{2}\ge \sqrt{p^2{q}^2}$$
$$\frac{1}{2}\ge pq$$
Multiplying by 2:
$$1\ge 2pq$$
So, $(p+q{)}^2=1+2pq\le 1+1=2$.
$$(p+q{)}^2\le 2⟹p+q\le \sqrt{2}$$
The maximum value is $\sqrt{2}$, occurring when $p=q=1/\sqrt{2}$.