Question

In any triangle ABC,$\sum \frac{{\sin }^{2}A+\sin A+1}{\sin A}$ is always greater than

Answer 1

In a triangle $ABC$

$\sum \frac{{\sin }^{2}A+\sin A+1}{\sin A}$

$=\frac{{\sin }^{2}A+\sin A+1}{\sin A}+\frac{{\sin }^{2}B+\sin B+1}{\sin B}+\frac{{\sin }^{2}C+\sin C+1}{\sin C}$

$=\frac{{\sin }^{2}A}{\sin A}+\frac{\sin A}{\sin A}+\frac{1}{\sin A}+\frac{{\sin }^{2}B}{\sin B}+\frac{\sin B}{\sin B}+\frac{1}{\sin B}+\frac{{\sin }^{2}C}{\sin C}+\frac{\sin C}{\sin C}+\frac{1}{\sin C}$

$=\sin A+1+\frac{1}{\sin A}+\sin B+1+\frac{1}{\sin B}+\sin C+1+\frac{1}{\sin C}$

$=\sin A+\frac{1}{\sin A}+\sin B+\frac{1}{\sin B}+\sin C+\frac{1}{\sin C}+3$

As $\sin \theta$ in any triangle is always greater than $0$.

$\therefore \sin \theta +\frac{1}{\sin \theta }\ge 2$

$\therefore \sin A+\frac{1}{\sin A}+\sin B+\frac{1}{\sin B}+\sin C+\frac{1}{\sin C}\ge 6$

$\Rightarrow \sin A+\frac{1}{\sin A}+\sin B+\frac{1}{\sin B}+\sin C+\frac{1}{\sin C}+3\ge 6+3$

$\Rightarrow \sin A+\frac{1}{\sin A}+\sin B+\frac{1}{\sin B}+\sin C+\frac{1}{\sin C}+3\ge 9$

Thus,$\sum \frac{{\sin }^{2}A+\sin A+1}{\sin A}\ge 9$