Trigonometry (5)

Call $\dfrac{A}2 = x , \dfrac{B}2 = y , \dfrac{C}2 = z$

$x+y+z = 90$

$\tan(x+y) = \tan(90-z)$

$\dfrac{\tan(x) + tan(y)}{1-\tan(x)\tan(y)} = \dfrac1{\tan(z)}$

$\tan(x)\tan(y)+\tan(x)\tan(z)+\tan(y)\tan(z) = 1$

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By AM-GM Inequality ,

$\tan^2(x) + \tan^2(y) + \tan^2(z) \geq \tan(x)\tan(y)+\tan(x)\tan(z)+\tan(y)\tan(z) = 1$

Equality holds when $\tan(x) = \tan(y) = \tan(z) \Rightarrow A=B=C=60^{\circ}$

Apply Jensen's inequality on the convex function $f(x)=tan^2(x)$ in (0,π/2) taking the positive multipliers equal to 1/3. Answer directly follows.

What I did was we kniw minimum occurs when A=B=C. Si A=B=C=60 (equality occurs here)

So putting values we get 1