Minimum area of a triangle In triangle inscribed circle with radius $r = 1$ and one of it sides $a=3$. Find the minimum area of triangle? Ans = 5.4
My reasonings:

$BC = a$, $AC = b$, $AB = c$
$AD=AF=x$
$FC=CE=y$
$BD=BE=z$
$a=z+y$, $b=x+y$, $c=x+z$
The radius of the incircle is $$r =\frac{A_{ABC}}{s}$$ where $s = \frac{a+b+c}{2} =  x + y+ z$. By condition $z+y=3$ so $s=x+3$
By Heron's formula, the area of the triangle is $A=\sqrt{s(s-a)(s-b)(s-c)}$
In other side $A=sr = x+3$.
What is next? I think that I should get a fucntion for which I will can find a minimum, but I don't know how.

Trigonometric approach:
In your notation, 
$AD=AF=x$, $FC=CE=y$, $BD=BE=z$,
denote in addition
$\alpha=\angle OAF=\angle OAD$,
$~~~\beta=\angle OCF=\angle OCE$,
$~~~\gamma=\angle OBE=\angle OBD$.
So, if $r=1$, then
$x=\dfrac{1}{\tan\alpha}$,
$~~~y=\dfrac{1}{\tan\beta}$, 
$~~~z=\dfrac{1}{\tan\gamma}$. 
Then, as you said, 
$$
A=r(x+y+z)=r(x+3)=x+3.
$$
$$
x = \dfrac{1}{\tan(90^\circ - \beta-\gamma)} = \tan(\beta+\gamma) = \dfrac{\tan\beta+\tan\gamma}{1-\tan\beta\tan\gamma}.
$$
Dividing numerator and denominator by $(\tan\beta\tan\gamma)$, we get:
$$
x=\frac{z+y}{yz-1}=\frac{3}{yz-1}.
$$
$$
A=x+3=\frac{3}{yz-1}+3.
$$
Further thoughts - as in answer of ElThor.
