Question

In any $\Delta ABC,\sum \left(\frac{{\sin }^{2}A+\sin A+1}{\sin A}\right)$ is always greater than

• A. 9
• B. 3
• C. 27
• D. $\frac{1}{3}$

Answer 1

The correct option is A 9

Let $y=\sum \left(\frac{{\sin }^{2}A+\sin A+1}{\sin A}\right)=\sum [1+(\sin A+\frac{1}{\sin A})]$
$\Rightarrow y=[1+(\sin A+\frac{1}{\sin A})]+[1+(\sin B+\frac{1}{\sin B})]+[1+(\sin C+\frac{1}{\sin C})]$
Now since $A,B,C$ are angles of a triangle $0<A,B,C<\pi$
$\Rightarrow \sin A,\sin B,\sin C>0$
Thus using $A.M.\ge G.M.$
$(\sin A+\frac{1}{\sin A})>2,(\sin B+\frac{1}{\sin B})>2,(\sin C+\frac{1}{\sin C})>2$
Hence $y>(1+2)+(1+2)+(1+2)=9$