Question

In any triangle ABC, the inequality \[ \sin \frac{A}{2} \leq \frac{a}{2\sqrt{bc}} \] holds. Determine the correct expression.

• A. \(\frac{2a}{\sqrt{bc}}\)
• B. \(\frac{a}{2\sqrt{bc}}\)
• C. \(\frac{3a}{\sqrt{bc}}\)
• D. \(\frac{\sqrt{bc}}{2a}\)

Hint:

Use the half-angle formula for sine in terms of triangle sides and simplify using inequalities like AM-GM.

Answer 1

The Correct Option is B

Step 1: Use sine formula.
\[ \sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}, \quad s = \frac{a+b+c}{2} \]

Step 2: Apply inequality.
\[ (s-b)(s-c) \leq \frac{bc}{4} \Rightarrow \sin \frac{A}{2} \leq \frac{a}{2\sqrt{bc}} \]

Step 3: Express in terms of sides.
\[ \sin \frac{A}{2} = \sqrt{\frac{(\frac{a+b+c}{2}-b)(\frac{a+b+c}{2}-c)}{bc}} = \sqrt{\frac{(a+c-b)(a+b-c)}{4bc}} \]

Step 4: Apply AM-GM or known inequality.
\[ (a+c-b)(a+b-c) \le a^2 \Rightarrow \sin \frac{A}{2} \le \frac{a}{2\sqrt{bc}} \]

Step 5: Final conclusion.
\[ \boxed{\sin \frac{A}{2} \le \frac{a}{2\sqrt{bc}}} \]