BM01 2011/12 Question 6 Geometry Problem 

*Let $ABC$ be an acute-angled triangle. The feet of the altitudes from
$A,B$ and $C$ are $D, E$ and $F$ respectively. Prove that $DE +DF \le BC$
and determine the triangles for which equality holds.
The altitude from A is the line through A which is perpendicular to
BC. The foot of this altitude is the point D where it meets BC. The
other altitudes are similarly defined.


Thanks in advance for any contributions.

$\Delta DEF$ is the orthic triangle of $\Delta ABC$, and it is well-known that $DF=b\cos B$ and $DE=c\cos C$. Also using the well-known projection formulae, $a=b\cos C+c\cos B$. So the inequality is equivalent to:
$$b\cos B + c\cos C \le b\cos C+c\cos B$$
$$(b-c)(\cos B - \cos C) \le 0$$
This inequality is easily proved because $b\le c \implies \sin B \le \sin C \implies \cos B \ge \cos C$. Similar procedure applies for the $b \ge c$ case.
The equality clearly holds in isosceles triangles. $A,B,C$ and $a,b,c$ have usual meaning.
For info on orthic triangle, click here
