Triple Tan

Firstly, since $\alpha + \beta + \gamma = 180$ , then $\frac {\alpha + \beta + \gamma}{2} = 90$

Since $\tan(\frac {\alpha + \beta}{2}) = \cot\frac {\gamma}{2}$ , we know $\tan(\frac {\alpha}{2} +\frac {\beta}{2})* \tan\frac {\gamma}{2} = 1$ . we will get $\frac { \tan\frac {\alpha}{2} + \tan\frac {\beta}{2}}{1 - \tan\frac {\alpha}{2}\times\tan\frac {\beta}{2}} \times \tan\frac {\gamma}{2} = 1$ , which reduces to

$\tan\frac {\alpha}{2}\times\tan\frac {\beta}{2} + \tan\frac {\beta}{2}\times\tan\frac {\gamma}{2} + \tan\frac {\alpha}{2}\times\tan\frac {\gamma}{2}= 1$

Let $a = \tan\frac {\alpha}{2}\times\tan\frac {\beta}{2} +8$ , $b = \tan\frac {\gamma}{2}\times\tan\frac {\beta}{2} +8$ , $c = \tan\frac {\alpha}{2}\times\tan\frac {\gamma}{2} +8$ .

We know $a + b + c = 1 + 8 + 8 + 8 = 25$

Let X be $\sqrt {a} + \sqrt {b} + \sqrt {c}$ , $X^2 = a + b + c + 2\times\sqrt {ab} +2\times\sqrt {bc} + 2\times\sqrt {ac}$

Since we are looking for the maximum value, using AM GM inequality, we get : $X^2 \leq a + b + c + a + b +b + c + c + a = 75$ .

In this case, the maximum value is obtained when $a=b=c$ which reduced to $\alpha = \beta = \gamma =60 ^\circ$ .

We claim that if a,b,c are angles of a triangle, then tan(a/2)tan(b/2)+tan(b/2)tan(c/2)+tan(c/2)tan(a/2)=1.square root[(x+8)(y+8)]+square roof[(y+8)(z+8)]+square root[(z+8)(x+8)]}.

Using a+b+c=180, the addition formulas and the fact that tan x = cot(90-x), we have

tan(c/2)=cot((a+b)/2)=(1-(tan(a/2))(tan(b/2)))/(tan(a/2)+tan(b/2))---------(I)

By plugging in (I) into the left hand side of the claim to complete the proof of the claim.

Let x=tan(a/2), y=tan(b/2), z=tan(c/2).

The given formula will transform into (square root(x+8)+square root(y+8)+square root(z+8))^2 where x+ y+z=1.

Simplifying, the given formula will become 25+2{square root[(x+8)(y+8)]+square roof[(y+8)(z+8)]+square root[(z+8)(x+8)]}.

By Cauchy-Schwarz inequality and using the fact that x+y+z=1,

square root[(x+8)(y+8)]+square root[(y+8)(z+8)]+square root[(z+8)(x+8)] <= square root{[(x+8)+(y+8)+(z+8)][(y+8)+(z+8)+(x+8)]}=25.

The given formula is smaller than or equal to 25+2(25)=75.

Equality occurs when alpha=beta=gamma=60 degrees.

First, we show that $\tan \frac {\alpha}{2} \tan \frac {\beta}{2} + \tan \frac {\beta}{2} \tan \frac {\gamma}{2} + \tan \frac {\gamma}{2} \tan \frac {\alpha}{2} = 1$ . This follows because

$\begin{aligned} \tan \frac {\alpha}{2} \left( \tan \frac {\beta}{2} + \tan \frac {\gamma}{2}\right) &= \tan \frac {\alpha}{2} \left( \tan \frac {\beta}{2} + \tan \frac {\gamma}{2} \right) \times \frac { 1 - \tan \frac {\beta}{2} \tan \frac {\gamma}{2} } {1 - \tan \frac {\beta}{2} \tan \frac {\gamma}{2}} \\ &= \tan \frac {\alpha}{2} \tan \frac {\beta+\gamma}{2} \left(1- \tan \frac {\beta}{2} \tan \frac {\gamma}{2} \right) \\ &=1- \tan \frac {\beta}{2} \tan \frac {\gamma}{2} \\ \end{aligned}$

Hence, by AM-GM (or Cauchy-Schwarz), we know that $(\sqrt{A} + \sqrt{B} + \sqrt{C})^2 \leq 3(A + B + C)$ , so the expression is less than or equal to $3( \tan \frac {\alpha}{2} \tan \frac {\beta}{2} +8 + \tan \frac {\beta}{2} \tan \frac {\gamma}{2} + 8+ \tan \frac {\gamma}{2} \tan \frac {\alpha}{2} + 8 ) = 75$ .

Equality holds when $A=B=C = \frac {2\pi}{3}$ .

By standard identities: $\sin\tfrac12\alpha = \sqrt{\frac{(s-b)(s-c)}{bc}} \qquad \qquad \cos\tfrac12\alpha = \sqrt{\frac{s(s-a)}{bc}}$ (and similarly), so that $\tan\tfrac12\alpha \; = \; \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and similarly. Thus $\tan\tfrac12\alpha\tan\tfrac12\beta \; = \; \frac{s-c}{s}$ and similarly. We are being asked to maximize (over all triangles) $\left(\sqrt{9 - \frac{2a}{a+b+c}} + \sqrt{9 - \frac{2b}{a+b+c}} + \sqrt{9 - \frac{2c}{a+b+c}}\right)^2$ This (by a simple application of Cauchy-Schwarz) has a maximum value of $75$ when $a=b=c$ .