According to http://www.artofproblemsolving.com/community/c6h422667, we have in general $[DEF]=\tfrac{1}{8}(a^2+b^2+c^2)+[ABC].$ When $\angle BAC=90^{\circ},$ we have $\frac{[DEF]}{[ABC]}=\frac{a^2+b^2+c^2}{8[ABC]}+1=\frac{2(b^2+c^2)}{8 \cdot \frac{1}{2}bc}=\frac{b^2+c^2}{2bc}+1.$ Thus, it immediately follows that $\tfrac{[DEF]}{[ABC]}>1$ and $\tfrac{[DEF]}{[ABC]} \ge 2$ is equivalent to $b^2+c^2 \ge 2bc,$ which is clearly true.