2015 Canada MO #2

 Let $ABC$ be an acute-angled triangle with altitudes $AD,BE,$ and $CF$. Let $H$ be the orthocentre, that is, the point where the altitudes meet. Prove that\[\frac{AB\cdot AC+BC\cdot CA+CA\cdot CB}{AH\cdot AD+BH\cdot BE+CH\cdot CF}\leq 2.\]

lol I know I just said I'll be posting geo less frequently but looking at the 2015 Canadian MO problems this one was way too tempting :P

By PoP, $AH\cdot AD=AF\cdot AB=AB\cdot AC\cos A$. Cycling through the sides and subbing, the inequality is equivalent to $\frac{ab+bc+ca}{ab\cos C+bc\cos A+ca\cos B}\leq2$. By LoC, $ab\cos C=\frac{a^2+b^2-c^2}{2}$. Subbing that in, the inequality becomes $\frac{ab+bc+ca}{\frac{a^2+b^2+c^2}{2}}\leq2$, which rearranges to $a^2+b^2+c^2\geq ab+bc+ca$, which we know is true. $\square$


Popular posts from this blog

1995 IMO #2

2014 IMO SL #C2

2015 IMO SL #A1