Lemma: Through the incenter $I$ of $\triangle{ABC}$ draw a line that meets the sides $AB$ and $AC$ at $P$ and $Q$, then: \[ \frac{AB}{AP} \cdot AC + \frac{AC}{AQ} \cdot AB = AB+BC+AC \] Proof of the lemma: Consider the general case: $M$ is any point on side $BC$ and $PQ$ is a line cutting AB, AM, AC at P, N, Q. Then: $\frac{AM}{AN}=\frac{S_{APMQ}}{\triangle{APQ}}=\frac{\triangle{APM}+\triangle{AQM}}{\triangle{PQA}}=\frac{\frac{AP}{AB}\triangle{ABM}+\frac{AQ}{AC}\triangle{ACM}}{\frac{AP\cdot AQ}{AB \cdot AC}}=$ $=\frac{AC}{AQ}\cdot \frac{BM}{BC}+\frac{AB}{AP}\cdot \frac{CM}{BC}$ If $N$ is the incentre then $\frac{AM}{AN}=\frac{AB+BC+CA}{AB+AC}$, $\frac{BM}{BC}=\frac{AB}{AB+AC}$ and $\frac{CM}{BC}=\frac{AC}{AC+AB}$. Plug them in we get: \[ \frac{AB}{AP} \cdot AC + \frac{AC}{AQ} \cdot AB = AB+BC+AC \] Back to the problem Let $I_1$ and $I_2$ be the areas of $\triangle{ABD}$ and $\triangle{ACD}$ and $E$ be the intersection of $KL$ and $AD$. Thus apply our formula in the two triangles we get: \[ \frac{AD}{AE} \cdot AB + \frac{AB}{AK} \cdot AD = AB+BD+AD \] and \[ \frac{AD}{AE} \cdot AC + \frac{AC}{AL} \cdot AD = AC+CD+AD \] Cancel out the term $\frac{AD}{AE}$, we get: \[ \frac{AB+BD+AD-\frac{AB}{AK} \cdot AD }{AC+CD+AD- \frac{AC}{AL} \cdot AD }=\frac{AB}{AC} \] \[ AB \cdot CD + AB \cdot AD - \frac{AB \cdot AC \cdot AD}{AL}=AC \cdot BD+ AC \cdot AD -\frac{AB \cdot AC \cdot AD}{AK} \] \[ AB+AB \cdot \frac{CD}{AD}-\frac{AB \cdot AC}{AL}=AC+ AC \cdot \frac{BD}{AD} - \frac{AB \cdot AC}{AK} \] \[ AB+AC - \frac{AB \cdot AC}{AL}=AB+AC - \frac{AB \cdot AC}{AK} \] \[ \frac{AB \cdot AC}{AK} = \frac{AB \cdot AC}{AL} \] So we conclude $AK=AL$. Hence $\angle{AKI_1}=45^o=\angle{ADI_1}$ and $\angle{ALI_2}=45^o=\angle{ADI_2}$, thus $\triangle{AK_1} \cong \triangle{ADI_1}$ and $\triangle{ALI_2} \cong \triangle{ADI_2}$. Thus $AK=AD=AL$. So the area ratio is: \[ \frac{E}{E_1}=\frac{AB \cdot AC}{AD^2} = \frac{BC}{AD} =\frac{BD+CD}{\sqrt{BD \cdot CD}}\geq 2 \]

I think we can be a bit quicker than that. Let X and Y be respective incenters of $ABD$ and $ACD$. Note that by spiral similarity taking $BDA$ to $ADC$, we have $XDY$ being a 45 degree rotation and dilation of both $BDA$ and $ADC$ (of course rotating in opposite directions). It follows then that $AK = AL$. Then $\angle AKX = 45 = \angle ADX \implies AK = AD$. Similarly, $AL = AD$. Then $\frac{[ABC]}{[AKL]}= \frac{AB \cdot AC}{AK \cdot AL}= \frac{AB \cdot AC}{AD^{2}}$. It's an easy finish from here; as shobber did, we can note that $BC \ge 2AD$, and so $\frac{AB \cdot AC}{AD^{2}}= \frac{BC}{AB}\cdot \frac{AB}{AD}= \frac{BC}{AD}\ge 2$.

Remark. Given are a triangle $ABC$ and a point $D\in [BC]$. Denote the incircles $C(I_{1},r_{1})$, $C(I_{2},r_{2})$ of the triangles $ABD$, $ACD$ respectively and the points $M\in AB\cap DI_{1}$, $N\in AC\cap DI_{2}$, $X\in AB\cap I_{1}I_{2}$, $Y\AC\cap I_{1}I_{2}$. Prove that $XY\parallel MN\Longleftrightarrow AX=AY$.

Remark by 28121941: This was problem 5 at IMO 1988, proposed by Greece. As (ABC) (area) is greater or equal 2 times (AKL), the minimal value of the quotient is 2. At Gazeta Matematica 1991(number 10), Neculai Roman published two generalizations of this problem: 1) Let ABC a triangle. Let k_1 the circle through A and B and tangent to AC; analogously, let k_2 the circle through A and C and tangent to AB. The second intersection of k_1 and k_2 is D. The line defined by the incenters of the triangles ABD and ACD meet the lines AB,AC in K and L, respectively. If S is the area of ABDC and T is the area of ACD, show that S is greater or equal to 2 times T. 2) Let ABC a triangle. with A>B, A>C. let D and D' two points of the segment BC such that angle CAD = angle ABC and angle BAD' = angle ACB. The line which join the incenters of ABD and ACD' intersect AB at K and AC at L. If S = area(ABC) and T = area(AKL), then S is greater or equal to 4T*(sin(A/2))^2.

Here's a thoughtless bary bash orl wrote: In a right-angled triangle $ ABC$ let $ AD$ be the altitude drawn to the hypotenuse and let the straight line joining the incentres of the triangles $ ABD, ACD$ intersect the sides $ AB, AC$ at the points $ K,L$ respectively. If $ E$ and $ E_1$ dnote the areas of triangles $ ABC$ and $ AKL$ respectively, show that \[ \frac {E}{E_1} \geq 2. \] Set $ABC$ as reference triangle.Let $P=AI_1\cap BC$ and $Q=AI_2\cap BC$.Then clearly we have \[P=(0:b:a-b)\implies I_1=(a(a-b):bc:c(a-b))\]\[Q=(0:a-c:c)\implies I_2=(a(a-c):b(a-c):bc)\]Now let $K=(x:y:0)$.Then we have \[\left|\begin{array}{ccc}x & y& 0 \\ a(a-b) & bc & c(a-b) \\ a(a-c) & b(a-c) & bc\end{array}\right|=0\implies K=(a-b:b:0)\]Similarly $L=(a-c:0:c)$.Therefore \[\dfrac{[AKL]}{[ABC]}=\left|\begin{array}{ccc}1 & 0 & 0 \\ 1-\tfrac{b}{a} & \tfrac{b}{a} & 0 \\ 1-\tfrac{c}{a} & 0 & \tfrac{c}{a}\end{array}\right| = \dfrac{bc}{a^2}=\dfrac{bc}{b^2+c^2}\leq \dfrac{1}{2} \; \square\]Thus we are done.