Let $O$ be the circumcenter and $R$ the circumradius of $\triangle ABC$. Now, $\angle A'B'C'=\angle A'AB+\angle C'CB=\frac{\alpha+\theta}{2} \\\Rightarrow \angle A'OC'=\alpha+\theta \Rightarrow \angle A'C'O=\angle C'A'O= 90-\frac{\alpha+\theta}{2}=\frac{\beta}{2}$ By the identity $a\cos B+b\cos A=c$ it follows that $A'C'=R\cos\frac{\beta}{2}+R\cos\frac{\beta}{2}=2R\cos\frac{\beta}{2}$. In a similar way we find $A'B'=2R\cos\frac{\theta}{2}, B'C'=2R\cos\frac{\alpha}{2} \\\Rightarrow (A'B'C')=\frac{2R\cos\frac{\beta}{2}\cdot 2R\cos\frac{\theta}{2}\cdot 2R\cos\frac{\alpha}{2}}{4R}$ Then $(A'B'C')\geq(ABC) \Leftrightarrow \frac{2R\cos\frac{\beta}{2}\cdot 2R\cos\frac{\theta}{2}\cdot 2R\cos\frac{\alpha}{2}}{4R}\geq \frac{abc}{4R} \\\Leftrightarrow 8R^3\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\theta}{2}\geq abc=8R^3\sin\alpha \sin\beta \sin\theta \\\Leftrightarrow \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\theta}{2}\geq 8\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\theta}{2}\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\theta}{2} \\\Leftrightarrow \frac{1}{8}\geq \sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\theta}{2}$ but the last inequality is very well-known, so we're done.

Let K, L, M be the intersections of the bisectors of the angles $\angle A, \angle B, \angle C$ with the circumcircle (O) of the triangle $\triangle ABC$. For example, the angle $\angle M$ of the triangle $\triangle KLM$ is equal to $\angle M = \angle LMK = \angle LMC + \angle CMK = \angle LBC + \angle CAK = \frac{\angle B + \angle A}{2}$ and similarly, $\angle K = \frac{\angle C + \angle B}{2},\ \ \ \angle L = \frac{\angle A + \angle C}{2}$ The areas of the triangles $\triangle ABC, \triangle KLM$ are equal to $S(\triangle ABC) = \frac{abs}{4R} = 2R^2 \sin \widehat A \sin \widehat B \sin \widehat C$ $S(\triangle KLM) = \frac{klm}{4R} = 2R^2 \sin \widehat K \sin \widehat L \sin \widehat M = 2R^2 \sin \frac{\widehat B + \widehat C}{2} \sin \frac{\widehat C + \widehat A}{2} \sin \frac{\widehat A + \widehat B}{2}$ where a = BC, b = CA, c = AB, k = LM, l = MK, m = KL and R is the common circumradius of these 2 triangles. Thus we have to show that $\sin \widehat A \sin \widehat B \sin \widehat C \le \sin \frac{\widehat B + \widehat C}{2} \sin \frac{\widehat C + \widehat A}{2} \sin \frac{\widehat A + \widehat B}{2}$ with the constraint $\angle A + \angle B + \angle C = \pi$. Substituting $\frac{\angle A + \angle B}{2} = \frac \pi 2 - \frac{\angle C}{2}$ and $\sin \frac{\widehat A + \widehat B}{2} = \cos \frac{\widehat C}{2}$, this becomes $\sin \widehat A \sin \widehat B \sin \widehat C \le \cos \frac{\widehat A}{2} \cos \frac{\widehat B}{2} \cos \frac{\widehat C}{2}$ $8 \sin \frac{\widehat A}{2} \sin \frac{\widehat B}{2} \sin \frac{\widehat C}{2} \le 1$ This inequality has been proved numerous times on this forum, either in this form or in the equivalent form $8 \sin \frac{\widehat A}{2} \sin \frac{\widehat B}{2} \sin \frac{\widehat C}{2} = 4 \left(\cos \frac{\widehat A - \widehat B}{2} - \cos \frac{\widehat A + \widehat B}{2} \right) \cos \frac{\widehat A + \widehat B}{2} =$ $= 2 \left(\cos \widehat A + \cos \widehat B\right) - 2 \left(1 + \cos (\widehat A + \widehat B)\right) \le 1$ $\cos \widehat A + \cos \widehat B + \cos \widehat C \le \frac 3 2$ For example, it can be reduced to $r \le \frac R 2$, where r, R are the triangle inradius and circumradius (see OH and IH) or proved using Lagrange multipliers (see Triangle inequality).

Let $H$ denote the orthocenter, and let $AH$ meet the circumcircle at $A''$. Define $B''$ and $C''$ similarly. Clearly $\angle A''CB=\angle A''AB=90-\angle B=\angle BCH$, and $A''H$ is perpendicular to $BC$. Hence $A''$ is the reflection of $H$ in $BC$ and therefore $\triangle A''CB\cong\triangle HCB$. Similarly $\triangle C''BA\cong\triangle B''AC$. Therefore the area of hexagon $AB''CA''BC''$ is 2 times the area of $ABC$. Let $B'C'$ meet $AA'$ at $X$. Then \[\angle B'XA=\angle B'C'A+\angle XAC'=\angle B'C'A+\angle XAB+\angle C''AB=\angle B'BA+\angle A'AB+\angle C'CB=\angle B/2+\angle A/2+\angle C/2+90.\] Hence $A'A$ is an altitude of $\triangle A'B'C'$. Hence the area of hexagon $AB'CA'BC'$ is twice of the area of triangle $A'B'C'$. However, $A'$ is the midpoint of the arc $BC$, so $[A'BC]\ge [A''BC]$. Similarly, $[AB'CA'BC']\ge [AB''CA''BC'']$. Hence $[A'B'C']\ge[ABC]$, as desired $\Box$.

Here is the sketch of another proof, partially thanks to RSM. Note that, if $A_1=AI\cap \odot(ABC);$ and if $A_2=BB_1\cap CC_1$ and if $B_1, B_2, C_1,C_2$ are defined similarly, then we get that $A_1B_1C_1$ is the medial triangle of $A_2B_2C_2.$ So we have to show that the medial triangle has the maximum area among all cevian triangles. Let $B_2C_2, C_2A_2$ and $A_2B_2$ be divided in the ratios $x:y, y:z, z:x$ by the points $A,B,C$ respectively. Then we see that $\frac{[A_2B_2C_2]}{[ABC]}=\frac{(x+y)(y+z)(z+x)}{4xyz}\geq 2$ from the AM-GM inequality. So, we are done. $\Box$

Nice

EDIT: Doesn't work, sorry for the spam. The boldened part is exactly where the proof goes wrong

Let $R$ denote circumradius. WLOG assume $R=\frac{1}2$, by extended law of sines, $x=\sin X$ for any side $x$ on circle, and $\angle X$ is angle subtended by side $x$ on circle. Some angle chase gives us $\angle A'=\angle \frac{1}2(B+C)$, symmetric values for $\angle B'$ and $\angle C'$. We know $$[ABC]=\frac{abc}{4R}=\frac{1}2 \left [\sin A \cdot \sin B \cdot \sin C \right]$$ Similarly $$[A'B'C']=\frac{1}2\left [\sin \frac{1}2(B+C) \cdot \sin \frac{1}2(C+A) \cdot \sin \frac{1}2(A+B) \right]$$ This reduces proving $$\sin \frac{1}2(B+C) \cdot \sin \frac{1}2(C+A) \cdot \sin \frac{1}2(A+B) \geq \sin A \cdot \sin B \cdot \sin C$$ Expanding values $\sin A=2 \sin \frac{A}2 \cos \frac{A}2$ and $\sin \frac{1}2 (B+C)=\cos \frac {A}2$ reduces the above statement to proving $$\sin \frac{A}2 \cdot \sin \frac {B}2 \cdot \sin \frac{C}2 \leq \frac {1}8$$ which is true by AM-GM.