Question

In a triangle $ABC$, the minimum value of $\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$ is equal to

• A. $3$
• B. $\frac{1}{3}$
• C. $1$
• D. $2$

Answer 1

The correct option is A $3$

$\begin{array}{c}Accordingtoquestion:\\ suppose,\\ x=\sqrt{b}+\sqrt{c}-\sqrt{a}\\ y=\sqrt{c}+\sqrt{a}-\sqrt{b}\\ and,\\ z=\sqrt{a}+\sqrt{b}-\sqrt{c}\\ Now,addingyandz:\\ y+z=\sqrt{c}+\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}-\sqrt{c}\\ y+z=2\sqrt{a}\\ \therefore \sqrt{a}=\frac{y+z}{2}\\ Again,addingzandx:\\ z+x=\sqrt{a}+\sqrt{b}-\sqrt{c}+\sqrt{b}+\sqrt{c}-\sqrt{a}\\ z+x=2\sqrt{b}\\ \therefore \sqrt{b}=\frac{z+x}{2}\\ and,addingxandy:\\ x+y=\sqrt{b}+\sqrt{c}-\sqrt{a}+\sqrt{c}+\sqrt{a}-\sqrt{b}\\ x+y=2\sqrt{c}\\ \therefore \sqrt{c}=\frac{x+y}{2}\\ substitutingfor\sqrt{a}andxin.\\ {\sum }^{\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}}\\ S=\sum \frac{\frac{y+z}{2}}{\frac{z+x}{2}+\frac{x+y}{2}-\frac{y+z}{2}}=\frac{y+z}{2}\left(\frac{2}{z+x+x+y-y-z}\right)=\frac{2}{2x}\left(\frac{y+z}{2}\right)=\frac{y+z}{2x}\\ S=\sum \frac{y+z}{2x}\\ =\frac{1}{2}\left[\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}\right]\\ =\frac{1}{2}\left[\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)\right]\\ =\frac{1}{2}\left(2+2+2\right)=3\\ So,thatthecorrectoptionisA.\end{array}$