Kanada 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, $$\begin{array}{rl}AH\cdot AD+BH\cdot BE+CH\cdot CF& =\frac{b^2+c^2-a^2}{2}+\frac{a^2+c^2-b^2}{2}+\frac{a^2+b^2-c^2}{2}\\ & =\frac{a^2+b^2+c^2}{2}.(1)\end{array}$$ 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., $△ABC$ is equilateral. $\square$

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