Minimum area of a triangle

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 $y = z = 1.5$

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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.

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.