Step-by-step Solution:

\[ A + B + C = 180^\circ \] From this, we can derive: \[ C = 180^\circ - (A + B) \] Expressing \(C/2\) in radians: \[ \frac{C}{2} = \frac{\pi}{2} - \frac{A + B}{2} \] Using the tangent formula: \[ \tan\left(\frac{C}{2}\right) = \tan\left(\frac{\pi}{2} - \frac{A + B}{2}\right) = \cot\left(\frac{A + B}{2}\right) \] The cotangent can be expressed as: \[ \cot\left(\frac{A + B}{2}\right) = \frac{1 - \tan\left(\frac{A}{2}\right)\tan\left(\frac{B}{2}\right)}{\tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right)} \] Let: \[ a = \tan\left(\frac{A}{2}\right), \; b = \tan\left(\frac{B}{2}\right), \; c = \tan\left(\frac{C}{2}\right) \] Since all values are positive, the constraint becomes: \[ c = \frac{1 - ab}{a + b} \] This leads to the equivalence: \[ ab + bc + ca = 1 \] \[\] Key Derivation From the above, we know: \[ a^2 + b^2 + c^2 \geq ab + bc + ca \] Using \(ab + bc + ca = 1\), we derive: \[ a^2 + b^2 + c^2 \geq 1 \] Substituting back: \[ \tan^2\left(\frac{A}{2}\right) + \tan^2\left(\frac{B}{2}\right) + \tan^2\left(\frac{C}{2}\right) \geq 1 \] \[\] Final Answer: \[ \boxed{\geq 1} \]