Mathematics - Properties of Triangle Question with Solution | TestHub

$$\begin{array}{rl}LHS& =\frac{1}{2}[b+b\cos A+a+a\cos B]\\ & =\frac{1}{2}(a+b+c)\\ & =\frac{3c}{2}\\ \Rightarrow 2c& =a+b\end{array}$$

Then,

$$\frac{a+c}{2c-a}+\frac{b+c}{2c-b}=\frac{a+c}{b}+\frac{b+c}{a}=\frac{a}{b}+\frac{b}{a}+\frac{c}{b}+\frac{c}{a}\ge 4{\left(\frac{c^2}{ab}\right)}^{1/4}\ge 4$$

(Since $\frac{a+b}{2}\ge \sqrt{ab}\Rightarrow c^2\ge ab\Rightarrow \frac{c^2}{ab}\ge 1$)