$(H_AH_BH_C)\le (ABC)\leftrightarrow (H_AH_B)(H_BH_C)(H_CH_A)\le abc\ (1)$ because both triangles are inscribed in the same circle. However, it is easy to show that $H_CH_A=2b\cdot \cos B,$ $H_AH_B=2c\cdot \cos C,$ and $H_BH_C=2a\cdot \cos A.$ Thus: $(H_AH_B)(H_BH_C)(H_CH_A)=8abc\cdot\cos A\cdot\cos B\cdot\cos C\le abc,$ which is $(1)$. The last inequality is well known. Equality occurs when the triangle is equilateral.

solution for the"well known" inequality

Attachments:

Slightly different solution. Assume that $AH_A \cap BC =D$, $BH_b \cap AC =E$ and $CH_C \cap AB =F$. Easy angle chasing reveals that $\triangle DEF \sim H_AH_BH_C$ and they are also homothetic with homothety centre at point $H$. Since $\frac{HD}{HH_A} =\frac{1}{2} $, we deduce that: $$ 4S(\triangle DEF) = S(\triangle H_AH_BH_C) $$Problem reduces to show that: $$ S(\triangle BFD) + S(\triangle CDE) + S(\triangle AFE) \ge \frac{3}{4} S(\triangle ABC) \quad (1) $$Observe that $\triangle AFE \sim \triangle ACB$ and that $\frac{AF}{AC} = \cos A$. Thus (1) becomes: \begin{align*} \cos^2 A + \cos^2 B + \cos^2 C \ge \frac{3}{4} \\ \frac{9}{4} \ge \sin^2 A + \sin^2 B +\sin^2 C \\ 9R^2 \ge a^2+b^2+c^2 \end{align*}But this is well- known that $OH^2 =9R^2 -(a^2+b^2 +c^2) $ (this can be also easily proved using complex numbers with $\odot(ABC)$ as unit circle ). This ends the proof.

Schur's inequality in the geometry problem, yes please. Claim. $(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^2+b^2+c^2)\leq a^2b^2c^2$. Proof. After the expansion, we need $$\sum_{sym} a^2b^4\leq 3a^2b^2c^2+a^6+b^6+c^6,$$but this is true by Schur's inequality, equality holds iff $a=b=c$. Let $D,E,F$ be foot from $A,B,C$ to $BC,AC,AB$, respectively. Firstly, note that $S_{H_AH_BH_C}=4S_{DEF}$ by homothety, hence we want $4S_{DEF}\leq S_{ABC}$. Now, we use barycentric coordinates. Let $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Thus, \begin{align*} D&=\left(0,\frac{a^2+b^2-c^2}{2a^2},\frac{a^2+c^2-b^2}{2a^2}\right),\\E&=\left(\frac{b^2+a^2-c^2}{2b^2},0,\frac{b^2+c^2-a^2}{2b^2}\right),\\F&=\left(\frac{c^2+a^2-b^2}{2c^2},\frac{c^2+b^2-a^2}{2c^2},0\right).\end{align*}We need to show that $$\begin{vmatrix} 0 & \frac{a^2+b^2-c^2}{2a^2} & \frac{a^2+c^2-b^2}{2a^2}\\ \frac{b^2+a^2-c^2}{2b^2} & 0 & \frac{b^2+c^2-a^2}{2b^2}\\ \frac{c^2+a^2-b^2}{2c^2} & \frac{c^2+b^2-a^2}{2c^2} & 0 \end{vmatrix}\leq \frac{1}{4},$$which is equivalent to $(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^2+b^2+c^2)\leq a^2b^2c^2$. We are done by the claim.

Complex bash is pretty clean as well... maybe i'll post a sol tomorrow if I'm in the mood