Question:

<p>Given <span class="math">A + B = \frac{\pi}{3}</span>, we have <span class="math">\tan(A + B) = \tan\frac{\pi}{3} = \sqrt{3}</span>.</p><p>Using the addition formula: <span class="math">\frac{\tan A + \tan B}{1 - \tan A \tan B} = \sqrt{3}</span>.</p><p>Let <span class="math">p = \tan A \tan B</span>. Then <span class="math">\tan A + \tan B = \sqrt{3}(1 - p)</span>.</p><p>By AM-GM inequality: <span class="math">\tan A + \tan B \geq 2\sqrt{\tan A \tan B}</span>.</p><p>So <span class="math">\sqrt{3}(1 - p) \geq 2\sqrt{p}</span>.</p><p>Squaring: <span class="math">3(1-p)^2 \geq 4p</span>, which gives <span class="math">3p^2 - 10p + 3 \geq 0</span>.</p><p>This factors as <span class="math">(3p - 1)(p - 3) \geq 0</span>, so <span class="math">p \leq \frac{1}{3}</span> or <span class="math">p \geq 3</span>.</p><p>Since <span class="math">A, B > 0</span> and <span class="math">A + B = \frac{\pi}{3} < \frac{\pi}{2}</span>, we must have <span class="math">p \leq \frac{1}{3}</span>.</p><p>Maximum value is <span class="math">\frac{1}{3}</span>, achieved when <span class="math">A = B = \frac{\pi}{6}</span>.</p>