Prove that in an acute angled triangle `ABC , sec A+sec B +sec Cge 6`.

Explanation

Triangle ABC is acute angled.
`:. 0 lt A, B, C gt pi//2`
So, consider the graph of function y = sec x for x `in (0, pi//2)`
The graph of function is concave upward.
Now, consider three points on the graph P (A sec, A), Q (B, sec B) and R(C, sec C)
Join these points to get triangle PQR

Now, centroid of the triangle PQR is
`G ((A + B + C)/(3), (sec A + sec B + sec C)/(3))`
Through point G, draw a line parallel to y-axis meeting graph at point D and x-axis at point E.
The abscissa of point D is `(A + B + C)/(3)`
So, coordinates of point D are `((A + B + C)/(3), sec ((A + B + C)/(3)))`
Now, clearly `GE ge DE`
`:. (sec A + sec B + sec C)/(3) ge sec (pi)/(3)`
`implies sec A + sec B + sec C ge 6`