Find the least value of `secA+secB+secC` in an acute angled triangle.

Explanation

Triangle ABC is acute angled.
`therefore 0ltA,B,Cltpi//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,secA),Q(B,secB)`
and `R(C,secC)`.
Join these points to get triangle PQR.
Centroid of the triangle PQR is
`G((A+B+C)/(3),(secA+secB+secC)/(3))`
Through point G, draw a line parallel to y-axis meeting graph to 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)))`
Clearly, `GEgeDE`
`therefore (secA+secB+secC)/(3)gesec((A+B+C)/(3))`
`rArr(secA+secB+secC)/(3)gesec""(pi)/(3)`
`rArr secA+secB+secCge6`