Question

If `A+B+C=pi,` prove that `tan^2A/2+tan^2B/2+tan^2C/2geq1.`

Answer 1

`(tan (A/2) - tan(B/2))^2 + (tan(B/2) - tan(C/2))^2 +(tan(C/2) - tan(A/2))^2 ge 0`
`=>2(tan^2(A/2)+tan^2(B/2)+tan^2(C/2)) - 2(tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)) ge 0->(1)`
Now,
`tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)`
`=tan(A/2)[tan(B/2)+tan(C/2)]+tan(B/2)tan(C/2)`
`=tan(A/2)[tan((B/2)+(C/2))(1-tan(B/2)tan(C/2)]+tan(B/2)tan(C/2)`
`=tan(A/2)[tan((B+C)/2)(1-tan(B/2)tan(C/2)]+tan(B/2)tan(C/2)`
`=tan(A/2)[tan((pi-A)/2)(1-tan(B/2)tan(C/2)]+tan(B/2)tan(C/2)`
`=tan(A/2)[tan(pi/2-A/2)(1-tan(B/2)tan(C/2)]+tan(B/2)tan(C/2)`
`=tan(A/2)cot(A/2)(1-tan(B/2)tan(C/2))+tan(B/2)tan(C/2)`
`=1-tan(B/2)tan(C/2)+tan(B/2)tan(C/2)`
`=1`
`=>tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2) = 1->(2)`
From (1) and (2),
`=>2(tan^2(A/2)+tan^2(B/2)+tan^2(C/2)) - 2 ge 0`
`=>2(tan^2(A/2)+tan^2(B/2)+tan^2(C/2)) ge 2`
`=>tan^2(A/2)+tan^2(B/2)+tan^2(C/2) ge 1`