Question

Find the least value of $\sec A+\sec B+\sec C$ in acute angle triangle

Answer 1

In a acute angle triangle, $\sec A$, $\sec B$ and $\sec C$ are positive.

Now A.M.$\ge$H.M.

$\Rightarrow$ $\frac{\sec A+\sec B+\sec C}{3}\ge \frac{3}{\cos A+\cos B+\cos C}$

Let $\cos A+\cos B+\cos C=x$

$\Rightarrow$ $2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\frac{C}{2}=x$

$2\sin \frac{C}{2}\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\frac{C}{2}=x$

$2{\sin }^2\frac{C}{2}-2\sin \frac{C}{2}\cos \left(\frac{A-B}{2}\right)+x-1-=0$
This is quadratic in $\sin C/2$ which is real. So, discriminant $D\ge 0$

$4{\cos }^2\left(\frac{A-B}{2}\right)-4\times 2(x-1)\ge 0$

$\Rightarrow$ $2(x-1)\le {\cos }^2\left(\frac{A-B}{2}\right)$

$2(x-1)\le 1$

$x\le 3/2$

Thus, $\cos A+\cos B+\cos C\le \frac{3}{2}$

$\frac{\sec A+\sec B+\sec C}{3}\ge 2$

$\sec A+\sec B+\sec C\ge 6$
Leat value=6