In triangle ABC , the minimum value of tan2A/2 + tan2B/2 + tan2C/2 is equal to
1P = expr.
P = tan2A/2 - 1 + tan2B/2 - 1 + tan2C/2 - 1 +3
P = -(2tanA/2tanA) -(2tanB/2tanB) -(2tanC/2tanC) + 3
Again A+B+C = Î
Therefore .... tanA/2 = cot(B+C)/2 ... (i)
Now for minimum value of P, ... P-3 must be maximised ....
Putting (i) in the expr. and maximising might give the ans ....
1Let f ( x ) = tan 2 ( x ) ........... x E [ 0 , π2 )
Since " f " is concave in the said interval , then applying Jensen's inequality ,
tan 2 ( A2 ) + tan 2 ( B2 ) + tan 2 ( C2 ) ≥ 3 tan 2 ( A + B + C6 ) = 1 .
For more on this inequality , search Wiki or Targetiit itself .
39tan(A/2) = r1/s
tan(B/2) = r2/s
tan(C/2) = r3/s
where ri = ex-radii and s = semi-perimeter of the triangle in question. Would this help in minimizing the expression?
E = 1s²(r1² + r2² + r3²)
39@Ricky : Is there any other way besides Jensen's inequality?
1Continuing after Pritish Bhaiyya ' s post -
r 1 r 2 + r 1 r 3 + r 3 r 2 = s 2
By AM - GM , s 2 ≥ 3 3√ r 1 r 2 r 3
But , r1 2 + r2 2 + r3 2 ≥ 3 3√ r 1 r 2 r 3 ( again by AM - GM )
Hence , E ( in Pritish Bhaiyya " s post ) ≥ s 2s 2 = 1
Equality - Equilateral Triangle : )
21or use this
if A + B + C = pi
tan (A/2) tan (B/2) + tan(B/2) tan(C/2) + tan (C/2) tan (A/2) = 1
so tan2(A/2) + tan 2(B/2) + tan2(C/2) ≥tan (A/2) tan (B/2) + tan(B/2) tan(C/2) + tan (C/2) tan (A/2)
or tan2(A/2) + tan 2(B/2) + tan2(C/2 ≥1
