Question

The least value of $\sec A+\sec B+\sec C$ in an acute angle triangle is

• A. $6$
• B. $4$
• C. $2$
• D. $0$

Answer 1

The correct option is A $6$

As all the angles are acute so, $\sec A,\sec B,\sec C$ are positive.

Now, $A.M.\ge H.M.$
$\Rightarrow \frac{\sec A+\sec B+\sec C}{3}\ge \frac{3}{\cos A+\cos B+\cos C}$

But in $△ABC,$ $\cos A+\cos B+\cos C\le \frac{3}{2}$
$\Rightarrow \frac{\sec A+\sec B+\sec C}{3}\ge 2\Rightarrow \sec A+\sec B+\sec C\ge 6$