In a triangle ABC prove that 2r<=(acotA +bcotB+ccotC)/3<=R.

First consider : -

( a cotA + b cotB + c cotC ) / 3 R

             2R (cosA + cosB + cosC ) / 3 R

cosA + cosB + cosC 3 / 2

2 cos(A+B)/2 cos(A-B)/2 + 1 - 2 sin²C/2 3 / 2

- 2 sin²C/2 + 2 (sinC/2)cos(A - B)/2 - 1 / 2 0

This a quadratic equation in sinC/2 withnegative coefficient of  sin²C/2

D = b2 - 4ac

= 4 cos²(A - B)/2 - 4

since cos is always less than 0 therefore  the eqauation is always valid with a critical point

Therefore,     

( a cotA + b cotB + c cotC ) / 3 R

Now, 2r   2R (cosA + cosB + cosc)

since r = 4RsinA/2sinB/2SinC/2

or to prove that

4RsinA/2sinB/2SinC/2 R (cosA + cosB + cosC ) / 3

On solving,

Consider s-b, s-c and use AM GM

(s-b + s-c ) / 2 (( s-b )(s-c ))^(1/2) / bc

a / 2(bc)^(1/2) sinA/2

similiarly for sinB/2 and sinC/2

therefore,

sinA/2sinB/2SinC/2 1/8

let sinA/2sinB/2SinC/2 = k

To prove 4k ( 4k + 1) / 3

or,   12k 4k + 1

or,         k 1/8 which is always possible.