Question

In $△ABC$, $b{\cos }^2\frac{A}{2}+a{\cos }^2\frac{B}{2}=\frac{3c}{2}$ then the minimum value of $\frac{a+c}{2c-a}+\frac{b+c}{2c-b}$ is

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is B $4$

We have

$b{\cos }^2\frac{A}{2}+a{\cos }^2\frac{B}{2}=\frac{3c}{2}$

$b\left(1+\cos A\right)+a\left(1+\cos B\right)=3c$

$b+a+c=3c\because c=b\cos A+a\cos B$

$b+a=2c$

$\frac{a+c}{2c-a}+\frac{b+c}{2c-b}=?$

$\frac{a+c}{b+a-a}+\frac{b+c}{b+a-b}$

$\frac{a\left(a+c\right)+b\left(b+c\right)}{ab}$

$=\frac{a^2+b^2+c\left(a+b\right)}{ab}$

$=\frac{a^2+b^2+c\left(2c\right)}{ab}$

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

$=\frac{6c^{2}2ab}{ab}$

$E=\frac{6c^2}{ab}-2⩾4$

$4c^2=a^2+b^2+2ab$

$\frac{a{c}^2}{ab}=2+\frac{a^2}{ab}+\frac{b^2}{ab}\because Am⩾Gm$

$\frac{4c^2}{ab}=2+\frac{a}{b}+\frac{b}{a}\frac{\frac{a}{b}+\frac{b}{a}}{2}⩾\sqrt{\frac{a}{b}+\frac{b}{a}}$

$\frac{4c^2}{ab}⩾4$

$\frac{c^2}{ab}⩾4$

Thus the required value is $4$

Hence, the option $\left(B\right)$ is the correct answer.