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Question

Show tan2(A/2)+tan2(B/2)+tan2(C/2)1 when A+B+C=π Hence the minimum value of tan2(A/2) is 1.

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Solution

We know that when A+B+C=π, then
tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)=1 .(1)
or yz+zx+xy=1, where x=tan(A/2) etc.
Now we know that (xy)2+(yz)2+(zx)20
or 2x22xy0
or x2xy or x21
xy=1
or tan2(A/2)+tan2(B/2)+tan2(C/2)1
Note: For another method see the chapter on inequalities where we have proved that x2+y2+z2xy+yz+zx=1x2+y2+z21.

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