Question:

<p><strong>Step 1:</strong> Apply Cauchy-Schwarz inequality (Lagrange's identity):</p><p><span class="math">$(a \tan \alpha + \sqrt{a^2 - 1} \tan \beta + \sqrt{a^2 + 1} \tan \gamma)^2 \leq (a^2 + a^2 - 1 + a^2 + 1)(\tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma)$</span></p><p><strong>Step 2:</strong> Substitute the constraint <span class="math">$a \tan \alpha + \sqrt{a^2 - 1} \tan \beta + \sqrt{a^2 + 1} \tan \gamma = 2a$</span>:</p><p><span class="math">$(2a)^2 \leq 3a^2(\tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma)$</span></p><p><strong>Step 3:</strong> Simplify:</p><p><span class="math">$4a^2 \leq 3a^2(\tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma)$</span></p><p><span class="math">$\tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma \geq \frac{4}{3}$</span></p><p><strong>Step 4:</strong> Multiply by 2727:</p><p><span class="math">$2727(\tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma) \geq 2727 \cdot \frac{4}{3} = 3636$</span></p><p>∴ Least value is <strong>3636</strong>.</p>