[ $[$ graphs and GMT on the coordinate plane $]$ Cauchy's Inequality $\quad]$

The graph of the linear function $y=k x+k+1$, where $k>0$, intersects the coordinate axes at points $A$ and $B$.

What is the smallest possible area of triangle $A B O$ ( $(O$ - the origin)?
The abscissa of the point of intersection of the graph with the $O X$ axis is $-\left(1+\frac{1}{k}\right)$. The ordinate of the point of intersection with the $O Y$ axis is $k+1$. Therefore,

$S_{A B O}=1 / 2 O A \cdot O B=1 / 2(k+1)(1+1 / k)=1 / 2(2+k+1 / k)$. The minimum value of the expression $k+1 / k$ is achieved when $k$ $=1$, hence, the smallest possible area of triangle $A B O$ is 2.