What is the minimum value of $(\tan^2)(A/2)+(\tan^2)(B/2)+(\tan^2)(C/2)$, where $A$, $B$ and $C$ are angles of a triangle?
I know that the sum of the angles is $\pi$, but I am unable to find the minimum value of the above expression.
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dknight
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Note that $A + B + C = \pi$. – Travis Willse Oct 14 '14 at 04:00
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I know that A+B+C=π – dknight Oct 14 '14 at 04:05
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3Possible duplicate of Trigonometric Functions And Identities Question . – MAN-MADE Aug 25 '17 at 07:32
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@MANMAID, you are right it is almost same as the one you stated. But do you even realise, it took you 3 years to search for that question, How was I supposed to find that question when I had to know the answer then. I can't keep waiting for 3 years just to find the answer of 1 question. – dknight Aug 28 '17 at 10:45
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Just another way: noting $\tan^2 \frac x 2$ is convex, by Jensen's inequality you have $$\tan^2\frac A 2 + \tan^2 \frac B 2 + \tan^2 \frac C 2 \ge 3\tan^2\frac{A+B+C}{2\cdot 3 } = 1$$
Macavity
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$\tan(C/2)=\tan(\pi/2-(A+B)/2)=\frac{1}{\tan((A+B)/2)}=\frac{1-\tan(A/2)\tan(B/2)}{\tan(A/2)+\tan(B/2)}$
Let $a=\tan(A/2), b=\tan(B/2), c=\tan(C/2)$, all positive, the constraint becomes
$$c=(1-ab)/(a+b)$$ which is equivalent to $ab+bc+ca=1$
I'll leave you to prove $a^2+b^2+c^2\geq ab+bc+ca$ and only equal when $a,b,c$ are all equal.
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By symmetry it occurs when triangle is equilateral, when $ A/2 = \pi/6 ; $
Minimum value $ 1/3 + 1/3 +1/3 =1.$
Narasimham
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