Question:
In a triangle \(ABC\),
\[
\left(
\tan\frac{A}{2}
\tan\frac{B}{2}
\tan\frac{C}{2}
\right)^2
\leq
\]
In a triangle \(ABC\),
\[
\left(
\tan\frac{A}{2}
\tan\frac{B}{2}
\tan\frac{C}{2}
\right)^2
\leq
\]
Show Hint
For a triangle,
\[
\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2}
\]
This identity is very useful for applying tangent sum formulas and AM-GM inequalities.
Updated On: Jun 22, 2026
- \(\dfrac{1}{27}\)
- \(\dfrac{1}{18}\)
- \(\dfrac{1}{9}\)
- \(\dfrac{1}{3}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use the angle sum property of a triangle.
In a triangle, \[ A+B+C=\pi \] Therefore, \[ \frac{A}{2}+\frac{B}{2}+\frac{C}{2} = \frac{\pi}{2} \]
Step 2: Use the standard identity.
For angles satisfying \[ x+y+z=\frac{\pi}{2}, \] we have \[ \tan x \tan y+\tan y \tan z+\tan z \tan x=1 \] Taking \[ x=\frac{A}{2}, \quad y=\frac{B}{2}, \quad z=\frac{C}{2}, \] we get \[ \tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1 \]
Step 3: Apply AM-GM inequality.
Let \[ p=\tan\frac{A}{2}\tan\frac{B}{2}, \] \[ q=\tan\frac{B}{2}\tan\frac{C}{2}, \] \[ r=\tan\frac{C}{2}\tan\frac{A}{2} \] Then, \[ p+q+r=1 \] By AM-GM inequality, \[ \frac{p+q+r}{3}\geq \sqrt[3]{pqr} \] Substituting, \[ \frac{1}{3}\geq \sqrt[3]{pqr} \] Cubing both sides, \[ \frac{1}{27}\geq pqr \] Now, \[ pqr = \left( \tan\frac{A}{2} \tan\frac{B}{2} \tan\frac{C}{2} \right)^2 \] Hence, \[ \left( \tan\frac{A}{2} \tan\frac{B}{2} \tan\frac{C}{2} \right)^2 \leq \frac{1}{27} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{1}{27}} \] which corresponds to option (1).
In a triangle, \[ A+B+C=\pi \] Therefore, \[ \frac{A}{2}+\frac{B}{2}+\frac{C}{2} = \frac{\pi}{2} \]
Step 2: Use the standard identity.
For angles satisfying \[ x+y+z=\frac{\pi}{2}, \] we have \[ \tan x \tan y+\tan y \tan z+\tan z \tan x=1 \] Taking \[ x=\frac{A}{2}, \quad y=\frac{B}{2}, \quad z=\frac{C}{2}, \] we get \[ \tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1 \]
Step 3: Apply AM-GM inequality.
Let \[ p=\tan\frac{A}{2}\tan\frac{B}{2}, \] \[ q=\tan\frac{B}{2}\tan\frac{C}{2}, \] \[ r=\tan\frac{C}{2}\tan\frac{A}{2} \] Then, \[ p+q+r=1 \] By AM-GM inequality, \[ \frac{p+q+r}{3}\geq \sqrt[3]{pqr} \] Substituting, \[ \frac{1}{3}\geq \sqrt[3]{pqr} \] Cubing both sides, \[ \frac{1}{27}\geq pqr \] Now, \[ pqr = \left( \tan\frac{A}{2} \tan\frac{B}{2} \tan\frac{C}{2} \right)^2 \] Hence, \[ \left( \tan\frac{A}{2} \tan\frac{B}{2} \tan\frac{C}{2} \right)^2 \leq \frac{1}{27} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{1}{27}} \] which corresponds to option (1).
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