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.

You were on the right track.
Using your notation, $A = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{(x + 3)xyz}$, but also $A = x + 3$. So
$$\begin{align}
\sqrt{(x + 3)xyz} &= x + 3, \\
xyz &= x + 3, \\
x &= \frac{3}{yz - 1}.
\end{align}$$
Substituting that into $A$ we get
$$A = \sqrt{\left(\frac{3}{yz - 1} + 3\right)\frac{3}{yz - 1}yz} = \sqrt{\frac{3yz}{yz - 1}\cdot\frac{3}{yz - 1}yz} = \frac{3yz}{yz - 1} = \frac{3}{yz - 1} + 3.$$
Finally, by AM-GM we get lower bound:
$$A = \frac{3}{yz - 1} + 3 \geqslant \frac{3}{\frac{(y + z)^2}{4} - 1} + 3 = \frac{3}{\frac{9}{4} - 1} + 3 = \frac{12}{5} + 3 = \frac{27}{5} = 5.4$$
Because it is AM-GM, equality is reached when $y = z = 1.5$

P.S. You may wonder, why for some values of $y$ and $z$ expression $\frac{3}{yz - 1}$ may become infinite or negative. Isn't it strange? Besides, I myself silently assumed that it is positive. And there's a reason for that.
Positive values of $\frac{3}{yz - 1}$ correspond to the situation you describe: a circle inscribe in a triangle.
When it becomes infinite two sides $AB$ and $AC$ become parallel.
And finally, when it's negative, your circle is no longer an incircle, it becomes excircle.
