Consider 3 circles of radii $r_1=r$, $r_2=2r$ and $r_3=3r$, where
$r=1/(2\sqrt{7\pi})$, so that their total area $s$ is
$$
s=\pi(r_1^2+r_2^2+r_3^2)=\frac12.
$$

<a href="http://en.wikipedia.org/wiki/Descartes%27_theorem">Descartes' theorem</a>
asserts that if three circles of radii $r_1$, $r_2$ and $r_3$
are pairwise externally tangent to each other and circumscribed by
the circle of radius $R$, then
$$
\frac1R=2\sqrt{\frac1{r_1r_2}+\frac1{r_2r_3}+\frac1{r_3r_1}}
-\biggl(\frac1{r_1}+\frac1{r_2}+\frac1{r_3}\biggr).
$$
In our case we find from this formula that
$$
R=\frac3{\sqrt{7\pi}},
$$
so that the area $S$ of the circumscribing circle is
$$
S=\pi R^2=\frac97>1.
$$
This implies that the three given circles of total area $1/2$
cannot be put inside a circle of total area 1 without intersections.

<strong>Edit.</strong> I have to agree that my geometric intuition is
too weak to notify an obvious non-applicability of Decartes' theorem.
As TonyK mentions in his comment, one can fit these three circles in
a circle of radius $r_2+r_3$, and the circle of radius $r_1$ fits
in one of the gaps, without touching the enclosing circle.

Without trying to correct the above solution I indicate another
choice: $r_1=0.99q$, $r_2=r_3=2q$, where $q$ is chosen in such a
way that the total area is again $1/2$. Decartes' theorem produces
the circumscribing circle of area $>1.008$. There is still an option
to put two large circles along a diameter of circle of total area 1
and see whether there is a room for the smaller one. This definitely
means that more geometry is involved...

<strong>Final edit</strong> (hopefully).
As roland-bacher mentions, the three circles again fit a circle
of radius $r_2+r_3$. After the two large equal circles are inscribed,
there is a room for another circle of radius $2r_2/3$ (well, one
can still have the space for the other of radius $2r_3/3$, but
I do not care of more than 3 circles inscribed). The corresponding
geometric picture involves the right triangles with sides $1$, $4/3$,
and $5/3$, a nice appearance of the Pythagorean triple $3^2+4^2=5^2$.

Is it true that this ($r_1=2/3$, $r_2=r_3=1$, total area $s=22\pi/9$)
is the worth case of inscribing 3 circles into the circle of
radius $R=2$ (area $S=4\pi$)? The quality of this inscription is
$S/s=18/11$. In other words, *can we replace areas $1/2$ and $1$
in the original problem by $11/18$ and $1$ respectively, if at least
three circles are inscribed*?

I have to apologize for my unsuccessful attempt. You still have
a chance to enjoy the beauty and difficulty of the original problem.
I thank TonyK and roland-bacher for pointing out my mistakes in geometry.