5. Let the incenter of $\triangle A B D$ be $M$, and the incenter of $\triangle A C D$ be $N$. Since $\triangle A D B \sim \triangle C D A$, note that $D M$ and $D N$ are two external angle bisectors from $D$, hence we have
$$\frac{n M}{D N}=\frac{B D}{A D},$$

and it is clear that $\angle M D N=90^{\circ}$. Therefore, $\triangle N M D$ is directly similar to $\triangle A B D$. Thus, the corresponding angles of the two triangles are equal, for example, the angle between $N M$ and $A B$ should equal the angle between $D M$ and $D A$, i.e.,
$$\angle L K A=\angle B D M=45^{\circ} \text {. }$$

This makes $\triangle A L K$ an isosceles right triangle. Also, since $\triangle A M K \cong \triangle A M D$, we have $A K=A D=A L$,
thus,
$$\begin{array}{l} S=\frac{1}{2} A B \cdot A C, \\ T=\frac{1}{2} A K \cdot A L=\frac{1}{2} A D^{2} \\ =\frac{1}{2} \frac{A B^{2} A C^{2}}{A B^{2}+A C^{2}} . \\ \end{array}$$

Therefore,
$$\frac{S}{2 T}=\frac{A B^{2}+A C^{2}}{2 A B \cdot A C^{2}} \geqslant 1 ;$$

which means $S \geqslant 2 T$. Proof complete.