Method 1: Let $A B=c, A C=b$, and $B C=a$ denote the three side lengths of the triangle.

As $\angle B F H=\angle B D H=90^{\circ}, F H D B$ is a cyclic quadrilateral. By the Power-of-a-Point Theorem, $A H \cdot A D=A F \cdot A B$. (We can derive this result in other ways: for example, see Method 2, below.)

Since $A F=A C \cdot \cos \angle A$, we have $A H \cdot A D=A C \cdot A B \cdot \cos \angle A=b c \cos \angle A$.

## Life Financial

Hence, by the Cosine Law, $\cos \angle A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$, which implies that $A H \cdot A D=\frac{b^{2}+c^{2}-a^{2}}{2}$.

By symmetry, we can show that $B H \cdot B E=\frac{a^{2}+c^{2}-b^{2}}{2}$ and $C H \cdot C F=\frac{a^{2}+b^{2}-c^{2}}{2}$.

$$\begin{aligned} A H \cdot A D+B H \cdot B E+C H \cdot C F & =\frac{b^{2}+c^{2}-a^{2}}{2}+\frac{a^{2}+c^{2}-b^{2}}{2}+\frac{a^{2}+b^{2}-c^{2}}{2} \\ & =\frac{a^{2}+b^{2}+c^{2}}{2} . \end{aligned}$$

Our desired inequality, $\frac{A B \cdot A C+B C \cdot B A+C A \cdot C B}{A H \cdot A D+B H \cdot B E+C H \cdot C F} \leq 2$, is equivalent to the inequality $\frac{c b+a c+b a}{\frac{a^{2}+b^{2}+c^{2}}{2}} \leq 2$, which simplifies to $2 a^{2}+2 b^{2}+2 c^{2} \geq 2 a b+2 b c+2 c a$.

But this last inequality is easy to prove, as it is equivalent to $(a-b)^{2}+(a-c)^{2}+(b-c)^{2} \geq 0$.

Therefore, we have established the desired inequality. The proof also shows that equality occurs if and only if $a=b=c$, i.e., $\triangle A B C$ is equilateral.

Method 2: Observe that

$$\frac{A E}{A H}=\cos (\angle H A E)=\frac{A D}{A C} \quad \text { and } \quad \frac{A F}{A H}=\cos (\angle H A F)=\frac{A D}{A B} \text {. }$$

It follows that

$$A C \cdot A E=A H \cdot A D=A B \cdot A F .$$

By symmetry, we similarly have

$$B C \cdot B D=B H \cdot B E=B F \cdot B A \quad \text { and } \quad C D \cdot C B=C H \cdot C F=C E \cdot C A \text {. }$$

Therefore

$$\begin{aligned} & 2(A H \cdot A D+B H \cdot B E+C H \cdot C F) \\ & \quad=A B(A F+B F)+A C(A E+C E)+B C(B D+C D) \\ & \quad=A B^{2}+A C^{2}+B C^{2} . \end{aligned}$$

This proves Equation (1) in Method 1. The rest of the proof is the same as the part of the proof of Method 1 that follows Equation (1).