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If A+B+C=pi, prove that tan^2(A/2)+tan^2...

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

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`(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)`
...
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