, 2015

Method 1: Let $AB = c$, $AC = b$, and $BC = a$ denote the three side lengths of the triangle.
As $\angle BFH = \angle BDH = 90^\circ$, $FHDB$ is a cyclic quadrilateral. By the Power-of-a-Point Theorem, $AH \cdot AD = AF \cdot AB$. (We can derive this result in other ways: for example, see Method 2, below.)
Since $AF = AC \cdot \cos \angle A$, we have $AH \cdot AD = AC \cdot AB \cdot \cos \angle A = bc \cos \angle A$.

By the Cosine Law, $\cos \angle A = \frac{b^2 + c^2 - a^2}{2bc}$, which implies that $AH \cdot AD = \frac{b^2 + c^2 - a^2}{2}$.
By symmetry, we can show that $BH \cdot BE = \frac{a^2 + c^2 - b^2}{2}$ and $CH \cdot CF = \frac{a^2 + b^2 - c^2}{2}$.
Hence,
$$\begin{aligned} AH \cdot AD + BH \cdot BE + CH \cdot CF &= \frac{b^2 + c^2 - a^2}{2} + \frac{a^2 + c^2 - b^2}{2} + \frac{a^2 + b^2 - c^2}{2} \\ &= \frac{a^2 + b^2 + c^2}{2}. \qquad (1) \end{aligned}$$
Our desired inequality, $\frac{AB \cdot AC + BC \cdot BA + CA \cdot CB}{AH \cdot AD + BH \cdot BE + CH \cdot CF} \le 2$, is equivalent to the inequality $\frac{cb + ac + ba}{\frac{a^2+b^2+c^2}{2}} \le 2$, which simplifies to $2a^2 + 2b^2 + 2c^2 \ge 2ab + 2bc + 2ca$.
But this last inequality is easy to prove, as it is equivalent to $(a-b)^2+(a-c)^2+(b-c)^2 \ge 0$.
Therefore, we have established the desired inequality. The proof also shows that equality occurs if and only if $a = b = c$, i.e., $\triangle ABC$ is equilateral. $\square$

Method 2: Observe that
$$\frac{AE}{AH} = \cos(\angle HAE) = \frac{AD}{AC} \quad \text{and} \quad \frac{AF}{AH} = \cos(\angle HAF) = \frac{AD}{AB}.$$
It follows that
$$AC \cdot AE = AH \cdot AD = AB \cdot AF.$$
By symmetry, we similarly have
$$BC \cdot BD = BH \cdot BE = BF \cdot BA \quad \text{and} \quad CD \cdot CB = CH \cdot CF = CE \cdot CA.$$
Therefore
$$\begin{aligned} & 2(AH \cdot AD + BH \cdot BE + CH \cdot CF) \\ &= AB(AF + BF) + AC(AE + CE) + BC(BD + CD) \\ &= AB^2 + AC^2 + BC^2. \end{aligned}$$
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). $\square$