Question

Let \(P\) be a real number and \(|P|\geq 2\). If \(A\), \(B\), \(C\) are variable angles such that \[ \sqrt{P^2-4}\tan A+P\tan B+\sqrt{P^2+4}\tan C=6P, \] then the minimum value of \[ \tan^2 A+\tan^2 B+\tan^2 C \] is

• A. \(6\)
• B. \(8\)
• C. \(12\)
• D. \(18\)

Hint:

For finding the minimum of \[ x^2+y^2+z^2 \] under a linear condition \[ ax+by+cz=k, \] use Cauchy-Schwarz inequality: \[ (a^2+b^2+c^2)(x^2+y^2+z^2)\geq k^2. \]

Answer 1

The Correct Option is C

Step 1: Substitute variables for tangent terms.
Let \[ x=\tan A,\quad y=\tan B,\quad z=\tan C \] Then the given condition becomes \[ \sqrt{P^2-4}x+Py+\sqrt{P^2+4}z=6P \] We need to find the minimum value of \[ x^2+y^2+z^2 \]

Step 2: Apply Cauchy-Schwarz inequality.
Using Cauchy-Schwarz inequality, \[ (a^2+b^2+c^2)(x^2+y^2+z^2)\geq (ax+by+cz)^2 \] Here, \[ a=\sqrt{P^2-4},\quad b=P,\quad c=\sqrt{P^2+4} \] So, \[ \left[(P^2-4)+P^2+(P^2+4)\right](x^2+y^2+z^2)\geq (6P)^2 \]

Step 3: Simplify the coefficient sum.
\[ (P^2-4)+P^2+(P^2+4)=3P^2 \] Thus, \[ 3P^2(x^2+y^2+z^2)\geq 36P^2 \]

Step 4: Divide by \(3P^2\).
Since \[ |P|\geq 2, \] we have \[ P^2\gt 0 \] Therefore, \[ x^2+y^2+z^2\geq \frac{36P^2}{3P^2} \] \[ x^2+y^2+z^2\geq 12 \]

Step 5: Check equality condition.
Equality in Cauchy-Schwarz holds when \[ \frac{x}{\sqrt{P^2-4}}=\frac{y}{P}=\frac{z}{\sqrt{P^2+4}} \] So the minimum value is attainable.

Step 6: Replace \(x\), \(y\), \(z\) by tangent terms.
Since \[ x=\tan A,\quad y=\tan B,\quad z=\tan C, \] we get \[ \tan^2 A+\tan^2 B+\tan^2 C\geq 12 \]

Step 7: Final conclusion.
Therefore, the minimum value is \[ \boxed{12} \]