If A+B+C=π, then find the minimum value of cot^2A+cot^2B+cot^2C.

We know that square of real numbers is always non-negative,

Hence

   (cotA-cotB)²+(cotB-cotC)²+(cotC-cotA)²≥0

⇒2(cot²A+cot²B+cot²C)-2(cotA·cotB+cotB·cotC+cotC·cotA)≥0

⇒(cot²A+cot²B+cot²C)≥(cotA·cotB+cotB·cotC+cotC·cotA)

We also know the trigonometric identity in a triangle that if Α+Β+C=π, then cotA·cotB+cotB·cotC+cotC·cotA=1,

∴,(cot²A+cot²B+cot²C)≥1.

Therefore, the minimum value of the given expression is 1.

 

The equality holds when

   (cot²A+cot²B+cot²C)=1

⇒(cotA-cotB)²+(cotB-cotC)²+(cotC-cotA)²=0       (Carrying out calculations identical to above, but in reverse order)

⇒cotA=cotB=cotC                                               (∵ squares are always non-negative, so if a sum of squares of real numbers is                                                                                 zero, then each term should be zero individually)

∵ cotA=cotB=cotC and Α+Β+C=π, 

∴ A=B=C=π/3

Hence, the equality holds for equilateral triangle.