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 $$ s=\pi(r_1^2+r_2^2+r_3^2)=\frac12. $$
Descartes' theorem 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.
Edit. 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...